## Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.2

Question 1.
Solve the following inequations graphically in a two-dimensional plane
(i) x ≤ -4
Solution:
Given, inequation is x ≤ -4
∴ corresponding equation is x = -4
It is a line parallel to Y-axis passing through the point A(-4, 0)
Origin test:
Substituting x = 0 in inequation, we get
0 ≤ -4 which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
So the points on the non-origin side of the line and points on the line satisfy the inequation
∴ all the points on the line and left of it satisfy the given inequation.
The shaded portion represents the solution set.

(ii) y ≥ 3
Solution:
Given, inequation is y ≥ 3
∴ corresponding equation is y = 3
It is a line parallel to X-axis passing through point A(0, 3)
Origin test:
Substituting y = 0 in inequation, we get
0 ≥ 3 which is false.
∴ Points on the origin side of the line do not satisfy the inequation
∴ Points on the non-origin side of the line satisfy the inequation.
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(iii) y ≤ -2x
Solution:
Given, inequation is y ≤ -2x
∴ corresponding equation is y = -2x
It is a line passing through origin O(0, 0).
To draw the line, we need one more point.
To find another point on the line, we can take any value of x,
say, x = 2.
∴ substituting x = 2 in y = -2x, we get
y = -2(2)
∴ y = -4
∴ another point on the line is A(2, -4)
Now, the origin test is not possible as the origin lies on the line y = -2x
So, choose a point which does not lie on the line say, (2, 1)
∴ substituting x = 2, y = 1 in inequation, we get
1 ≤ -2(2)
∴ 1 ≤ -4 which is false.
∴ the points on the side of the line y = -2x, where (2, 1) lies do not satisfy the inequation.
∴ all the points on the line y = -2x and on the opposite side of the line where (2, 1) lies, satisfy the inequation
The shaded portion represents the solution set.

(iv) y – 5x ≥ 0
Solution:
Given, inequation is y – 5x ≥ 0
∴ corresponding equation is y – 5x = 0
It is a line passing through the point O(0, 0)
To draw the line, we need one more point.
To find another point on the line,
we can take any value of x, say, x = 1.
Substituting x = 1 in y – 5x = 0, we get
y – 5(1) = 0
∴ y = 5
∴ Another point on the line is A(1, 5)
Now origin test is not possible as the origin lies on the line y = 5x
∴ choose a point that does not lie on the line, say (3, 2).
∴ substituting x = 3, y = 2 in inequation, we get
2 – 5(3) ≥ 0
∴ 2 – 10 ≥ 0
∴ -8 ≥ 0 which is false.
∴ the points on the side of line y = 5x where (3, 1) lies do not satisfy the inequation.
∴ the points on the line y = 5x and on the opposite of the line where (3, 2) lies, satisfy the inequation.
The shaded portion represents the solution set.

(v) x – y ≥ 0
Solution:
Given, inequation is x – y ≥ 0
∴ Corresponding equation is x – y = 0
It is a line passing through origin O(0, 0)
To draw the line we need one more point.
To find another point on the line, we can take any value of x,
Say, x = 2.
∴ substituting x = 2 in x – y = 0, we get
2 – y = 0
∴ y = 2
∴ another point on the line is A(2, 2)
Now origin test is not possible as the origin lies on the line y = x
∴ choose a point which not lie on the line say (3, 1)
∴ substituting x = 3, y = 1 in inequation, we get
3 – 1 ≥ 0
∴ 2 ≥ 0 which is true.
∴ all the points on line x – y = 0 and the points on the side where (3, 1) lies satisfy the inequation
The shaded portion represents the solution set.

(vi) 2x – y ≤ -2
Solution:
Given, inequation is 2x – y ≤ -2
∴ corresponding equation is 2x – y = -2
∴ $$\frac{2 x}{-2}-\frac{y}{-2}=\frac{-2}{-2}$$
∴ $$\frac{x}{-1}+\frac{y}{2}=1$$
∴ intersection of line with X-axis is A(-1, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
2(0) – (0) ≤ -2
∴ 0 ≤ -2
which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
∴ Points on the non-origin side of the line satisfy the inequation
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(vii) 4x + 5y ≤ 40
Solution:
Given, inequation is 4x + 5y ≤ 40
∴ Corresponding equation is 4x + 5y = 40
∴ $$\frac{4 x}{40}+\frac{5 y}{40}=\frac{40}{40}$$
∴ $$\frac{x}{10}+\frac{y}{8}=1$$
∴ Intersection of line with X-axis is A(10, 0)
Intersection of line with Y-axis is B(0, 8)
Origin test:
Substituting x = 0, y = 0 in the inequation, we get
4(0) + 5(0) ≤ 40
∴ 0 ≤ 40 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

(viii) $$\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y$$ ≤ 1
Solution:
Given, inequation is $$\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y$$ ≤ 1
∴ corresponding equation is $$\frac{x}{4}+\frac{y}{2}$$ = 1
∴ intersection of line with X-axis is A(4, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
$$\frac{1}{4}(0)+\frac{1}{2}(0)$$ ≤ 1
∴ 0 ≤ 1 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

Question 2.
Mr. Rajesh has ₹ 1,800 to spend on fruits for the meeting. Grapes cost ₹ 150 per kg. and peaches cost ₹ 200 per kg. Formulate and solve it graphically.
Solution:
Let x and y be the number of kgs. of grapes and peaches bought.
The cost of grapes is ₹ 150/- per kg, cost of peaches is ₹ 200/- per kg.
∴ cost of v kg of grapes is ₹ 150x
and the cost of y kg of peaches is ₹ 200y.
Mr. Rajesh has ₹ 1800 to spend on fruits.
∴ the total cost of grapes and peaches must be less than or equal to ₹ 1800.
∴ required inequation is 150x + 200y ≤ 1800
i.e., 3x + 4y ≤ 36 ……(i)
Since the number of kg of grapes and peaches can not be negative
∴ x ≥ 0, y ≥ 0
Now, corresponding equation is 3x + 4y = 36
∴ $$\frac{3 x}{36}+\frac{4 y}{36}=\frac{36}{36}$$
∴ $$\frac{x}{12}+\frac{y}{9}=1$$
∴ the intersection of the line with the X-axis is A(12, 0)
the intersection of the line with the Y-axis is B(0, 9)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
3(0) + 4(0) ≤ 36
∴ 0 ≤ 36 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is the points on the sides of the triangle OAB and in the interior of ∆OAB.
∴ the shaded portion represents the solution set.

Question 3.
The Diet of the sick person must contain at least 4000 units of vitamin. Each unit of food F1 contains 200 units of vitamin, whereas each unit of food F2 contains 100 units of vitamins. Write an inequation to fulfill a sick person’s requirements and represent the solution set graphically.
Solution:
Let the diet of the sick person contain, x units of food F1 and y units of food F2.
Since each unit of food F1 contains 200 units of vitamins.
∴ x units of food F1 contain 200x units of vitamins.
Also, each unit of food F2 contains 100 units of vitamins.
y units of food F2 contain 100y units of vitamins.
Now, Diet for a sick person must contain at least 4000 units of vitamins.
∴ he must take food F1 and F2 in such away that total vitamins must be greater than or equal to 4000.
∴ required inequation is 200x + 100y ≥ 4000
i.e., 2x + y ≥ 40
Also x and y cannot be negative.
∴ x ≥ 0, y ≥ 0
Corresponding equation is 2x + y = 40
∴ $$\frac{2 x}{40}+\frac{y}{40}=\frac{40}{40}$$
∴ $$\frac{x}{20}+\frac{y}{40}=1$$
∴ intersection of line with X-axis is A(20, 0)
intersection of line with Y-axis is B(0, 40)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
2(0) + (0) ≥ 40
∴ 0 ≥ 40 which is false
∴ all the points on the non origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is as shown in the figure.

## Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.1

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.1

Question 1.
Write the inequations that represent the interval and state whether the interval is bounded or unbounded:
(i) [-4, $$\frac{7}{3}$$]
Solution:
[-4, $$\frac{7}{3}$$]
Here, x takes values between -4 and $$\frac{7}{3}$$ including -4 and $$\frac{7}{3}$$
∴ the required inequation is -4 ≤ x ≤ $$\frac{7}{3}$$
∴ it is a bounded (closed) interval.

(ii) (0, 0.9]
Solution:
(0, 0.9]
Here, x takes values between 0 and 0.9, including 0.9 and excluding 0.
∴ the required inequation is 0 < x ≤ 0.9
∴ it is a bounded (semi-right closed) interval.

(iii) (-∞, ∞)
Solution:
(-∞, ∞)
Here, x takes values between -∞ and ∞
∴ the required inequation is -∞ < x < ∞
∴ it is an unbounded (open) interval.

(iv) [5, ∞)
Solution:
[5, ∞)
Here, x takes values between 5 and ∞ including 5.
∴ the required inequation is 5 ≤ x < ∞
∴ it is an unbounded (semi-left closed) interval.

(v) (-11, -2)
Solution:
(-11, -2)
Here, x takes values between -11 and -2
∴ the required inequation is -11 < x < -2
∴ it is a bounded (open) interval.

(vi) (-∞, 3)
Solution:
(-∞, 3)
Here, x takes values between -∞ and 3
∴ the required inequation is -∞ < x < 3
∴ it is an unbounded (open) interval.

Question 2.
Solve the following inequations
(i) 3x – 36 > 0
Solution:
3x – 36 > 0
Adding 36 both sides, we get
3x – 36 + 36 > 0 + 36
∴ 3x > 36
Dividing both sides by 3, we get
$$\frac{3 x}{3}>\frac{36}{3}$$
∴ x > 12
∴ x takes all real values more than 12
∴ Solution set = (12, ∞)

(ii) 7x – 25 ≤ -4
Solution:
7x – 25 ≤ -4
Adding 25 on both sides, we get
7x – 25 + 25 ≤ -4 + 25
∴ 7x ≤ 21
Dividing both sides by 7, we get
x ≤ 3
∴ x takes all real values less or equal to 3.
∴ Solution Set = (-∞, 3]

(iii) 0 < $$\frac{x-5}{4}$$ < 3
Solution:
0 < $$\frac{x-5}{4}$$ < 3
0 < x – 5 < 12
Adding 5 on both sides, we get
5 < x < 17
x takes all real values between 5 and 17.
∴ Solution set = (5, 17)

(iv) |7x – 4| < 10
Solution:
|7x – 4| < 10
-10 < 7x – 4 < 10 …….[|x| < k is same as -k < x < k]
Adding 4 on both sides, we get
-6 < 7x < 14
Dividing both sides by 7, we get
$$-\frac{6}{7}<x<\frac{14}{7}$$
∴ $$-\frac{6}{7}$$ < x < 2
∴ x takes all real values between $$-\frac{6}{7}$$ and 2.
∴ Solution set = ($$-\frac{6}{7}$$, 2)

Question 3.
Sketch the graph which represents the solution set for the following inequations:
(i) x > 5
Solution:
x > 5
Here, x takes all real values that are greater than 5.
∴ Solution set represents the unbounded (open) interval (5, ∞)
∴ the required graph of the solution set is as follows:

(ii) x ≥ 5
Solution:
x ≥ 5
Here, x takes all real values that are greater than or equal to 5
∴ Solution set represents the unbounded (semi-left closed) interval [5, ∞)
∴ the required graph of the solution set is as follows:

(iii) x < 3
Solution:
x < 3
Here, x takes all real values that are less than 3.
∴ Solution set represents the unbounded (open) interval (-∞, 3)
∴ the required graph of the solution set is as follows:

(iv) x ≤ 3
Solution:
x ≤ 3
Here, x takes all real values less than and including 3
∴ Solution set represents the unbounded (semi-right closed) interval (-∞, 3]
∴ the required graph of the solution set is as follows:

(v) -4 < x < 3
Solution:
-4 < x < 3
Here, x takes all real values between -4 and 3.
∴ Solution set represents the bounded (open) interval (-4, 3)
∴ the required graph of the solution set is as follows:

(vi) -2 ≤ x < 2.5
Solution:
-2 ≤ x < 2.5
Here, x takes all values between -2 and 2.5 including -2.
∴ Solution set represents the bounded (semi-left closed) interval [-2, 2.5)
∴ the required graph of the solution set is as follows.

(vii) -3 ≤ x ≤ 1
Solution:
-3 ≤ x ≤ 1
Here, x takes all real values between -3 and 1 including -3 and 1
∴ Solution set represents the bounded (closed) interval [-3, 1]
∴ the required graph of the solution set is as follows:

(viii) |x| < 4
Solution:
|x| < 4 ⇒ -4 < x < 4
Here, x takes all real values between -4 and 4.
∴ Solution set represents bounded (open) interval (-4, 4)
∴ the required graph of the solution set is as follows:

(ix) |x| ≥ 3.5
Solution:
|x| ≥ 3.5 ⇒ x ≥ 3.5 or x ≤ -3.5
Here, x takes values greater than or equal to 3.5 or it takes values less than or equal to -3.5
∴ Solution set represents the unbounded (semi-left closed) interval [3.5, ∞) or the unbounded (semi-right closed) interval (-∞, -3.5]
∴ x ∈ (-∞, -3.5] ∪ [3.5, ∞)
∴ the required graph of the solution set is as follows:

Question 4.
Solve the inequations:
(i) 5x + 7 > 4 – 2x
Solution:
5x + 7 > 4 – 2x
Adding 2x on both sides, we get
7x + 7 > 4
Subtracting 7 from both sides, we get
7x > -3
Dividing by 7 on both sides, we get
∴ x > $$-\frac{3}{7}$$
i.e., x takes all real values greater than $$-\frac{3}{7}$$
∴ the solution set is ($$-\frac{3}{7}$$, ∞)

(ii) 3x + 1 ≥ 6x – 4
Solution:
3x + 1 ≥ 6x – 4
Subtracting 3x from both sides, we get
1 ≥ 3x – 4
Adding 4 on both sides, we get
5 ≥ 3x
Dividing by 3 on both sides, we get
$$\frac{5}{3}$$ ≥ x
i.e., x ≤ $$\frac{5}{3}$$
i.e., x takes all real values less than or equal to $$\frac{5}{3}$$.
∴ the solution set is (-∞, $$\frac{5}{3}$$]

(iii) 4 – 2x < 3(3 – x)
Solution:
4 – 2x < 3(3 – x)
∴ 4 – 2x < 9 – 3x
Adding 3x on both sides, we get
4 + x < 9
Subtracting 4 from both sides, we get
x < 5
i.e., x takes all real values less than 5
∴ the solution set is (-∞, 5)

(iv) $$\frac{3}{4}$$x – 6 ≤ x – 7
Solution:
$$\frac{3}{4}$$x – 6 ≤ x – 7
Multiplying by 4 on both sides, we get
3x – 24 ≤ 4x – 28
Subtracting 3x from both sides, we get
-24 ≤ x – 28
Adding 28 on both the sides, we get
∴ 4 ≤ x i.e., x ≥ 4
i.e., x takes all real values greater or equal to 4.
∴ the solution set is [4, ∞)

(v) -8 ≤ -(3x – 5) < 13
Solution:
-8 < -(3x – 5) < 13 Multiplying by -1 throughout (so inequality sign changes) 8 ≥ 3x – 5 > -13
i.e., -13 < 3x – 5 ≤ 8
Adding 5 on both the sides, we get
-8 < 3x ≤ 13
Dividing, by 3 on both sides, we get
∴ $$-\frac{8}{3}$$ < x ≤ $$\frac{13}{3}$$
i.e., x takes all real values between $$-\frac{8}{3}$$ and $$\frac{13}{3}$$ including $$\frac{13}{3}$$.
∴ the solution set is $$\left(-\frac{8}{3}, \frac{13}{3}\right]$$

(vi) -1 < 3 – $$\frac{x}{5}$$ ≤ 1
Solution:
-1 < 3 – $$\frac{x}{5}$$ ≤ 1
Subtracting 3 from both sides, we get
-4 < –$$\frac{x}{5}$$ < -2 Multiplying by -1 throughout (so inequality sign changes) ∴ 4 > $$\frac{x}{5}$$ > 2
i.e., 2 < $$\frac{x}{5}$$ < 4
Multiplying by 5 on both sides, we get
10 < x < 20
i.e., x takes all real values between 10 and 20.
∴ the solution set is (10, 20)

(vii) 2|4 – 5x| ≥ 9
Solution:
2|4 – 5x | ≥ 9
∴ |4 – 5x| ≥ $$\frac{9}{2}$$
∴ 4 – 5x ≥ $$\frac{9}{2}$$ or 4 – 5x ≤ –$$\frac{9}{2}$$ ……[|x| ≥ a implies x ≤ -a or x ≥ a]
Subtracting 4 from both sides, we get
-5x ≥ $$\frac{1}{2}$$ or -5x ≤ $$\frac{-17}{2}$$
Divide by -5 (so inequality sign changes)
∴ x ≤ $$-\frac{1}{10}$$ or x ≥ $$\frac{17}{10}$$
∴ x takes all real values less than or equal to $$-\frac{1}{10}$$
or it takes all real values greater or equal to $$\frac{17}{10}$$.
∴ the solution set is (-∞, $$-\frac{1}{10}$$] or [$$\frac{17}{10}$$, ∞)

(viii) |2x + 7| ≤ 25
Solution:
|2x + 7| < 25
∴ -25 ≤ 2x + 7 ≤ 25 …..[|x| ≤ a implies -a ≤ x ≤ a]
Subtracting 7 from both sides, we get
-32 ≤ 2x ≤ 18
Dividing by 2 on both sides, we get
-16 ≤ x ≤ 9
∴ x can take all real values between -16 and 9 including -16 and 9.
∴ the solution set is [-16, 9]

(ix) 2|x + 3| > 1
Solution:
2|x + 3| > 1
Dividing by 2 on both sides, we get
|x + 3| > $$\frac{1}{2}$$
∴ x + 3 < –$$\frac{1}{2}$$ or x + 3 > $$\frac{1}{2}$$ …..[|x| > a implies x < -a or x > a]
Subtracting 3 from both sides, we get
x < – 3 – $$\frac{1}{2}$$ or x > -3 + $$\frac{1}{2}$$
∴ x < $$\frac{-7}{2}$$ or x > $$\frac{-5}{2}$$
∴ x can take all real values less $$\frac{-7}{2}$$ or it can take values greater than $$\frac{-5}{2}$$.
∴ Solution set is (-∞, $$\frac{-7}{2}$$) ∪ ($$\frac{-7}{2}$$, ∞)

(x) $$\frac{x+5}{x-3}$$ < 0
Solution:
$$\frac{x+5}{x-3}$$ < 0
Since $$\frac{a}{b}$$ < 0, when a > 0 and b < 0 or a < 0 and b > 0
∴ either x + 5 > 0 and x – 3 < 0
or x + 5 < 0 and x – 3 > 0
Case I:
x + 5 > 0 and x – 3 < 0 ∴ x > -5 and x < 3
∴ -5 < x < 3
∴ solution set = (-5, 3)
Case II:
x + 5 < 0 and x – 3 > 0
∴ x < -5 and x > 3
which is not possible
∴ solution set = Φ
∴ solution set of the given inequation is (-5, 3)

(xi) $$\frac{x-2}{x+5}$$ > 0
Solution:
$$\frac{x-2}{x+5}$$ > 0
Since $$\frac{a}{b}$$ > 0,
when a > 0 and b > 0 or a < 0 and b < 0 b
∴ either x – 2 > 0 and x + 5 > 0
or x – 2 < 0 and x + 5 < 0 Case I: x – 2 > 0 and x + 5 > 0
∴ x > 2 and x > -5
∴ x > 2
∴ solution set = (2, ∞)
Case II:
x – 2 < 0 and x + 5 < 0
∴ x < 2 and x < -5
∴ x < -5
∴ solution set = (-∞, -5)
∴ the solution set of the given inequation is (-∞, -5) ∪ (2, ∞)

Question 5.
Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x1, x2, x3 denote the marks in 1st, 2nd and 3rd unit test respectively. Then
$$\frac{x_{1}+x_{2}+x_{3}}{3}$$ ≥ 60
∴ $$\frac{70+75+x_{3}}{3}$$ ≥ 60
∴ 145 + x3 ≥ 3(60)
Subtracting 145 from both sides, we get
x3 ≥ 180 – 145
∴ x3 ≥ 35
Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.

Question 6.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade ‘A’ in the course.
Solution:
Let x1, x2, x3, x4, x5 denote the marks in five examinations. Then
$$\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}$$ ≥ 90
∴ $$\frac{87+92+94+95+x_{5}}{5}$$ ≥ 90
∴ 368 + x5 ≥ 450
Subtracting 368 from both sides, we get
∴ x5 ≥ 82
Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.

Question 7.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z,
Given that 2n – 1 < 10 and 2n + 1 < 10
∴ 2n < 11 and 2n < 9
∴ 2n < 9
∴ n < $$\frac{9}{2}$$ …..(i) Also, (2n – 1) + (2n + 1) > 11
∴ 4n > 11
∴ n > $$\frac{11}{4}$$ …….(ii)
From (i) and (ii)
$$\frac{11}{4}<n<\frac{9}{2}$$ Since, n is an integer,
∴ n = 3, 4
n = 3 gives 2n – 1 = 5, 2n + 1 = 7
and n = 4 gives 2n – 1 = 7, 2n + 1 = 9
∴ The pairs of positive consecutive integers are (5, 7) and (7, 9).

Question 8.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let 2n, 2n + 2 be two positive consecutive integers where n ≥ 1 ∈ Z.
Given that 2n > 5 and 2n + 2 > 5
∴ n > $$\frac{5}{2}$$ and 2n > 3
∴ n > $$\frac{5}{2}$$ and n > $$\frac{3}{2}$$
∴ n > $$\frac{5}{2}$$ ……(i)
Also (2n) + (2n + 2) < 23
∴ 4n + 2 < 23
∴ 4n < 21
∴ n < $$\frac{21}{4}$$ ……(ii)
From (i) and (ii)
$$\frac{5}{2}<n<\frac{21}{4}$$ and n is an integer.
∴ n = 3, 4, 5
n = 3 gives 2n = 6, 2n + 2 = 8
n = 4 gives 2n = 8, 2n + 2 = 10
n = 5 gives 2n = 10, 2n + 2 = 12
∴ The pairs of positive even consecutive integers are (6, 8) (8, 10), (10, 12)

Question 9.
The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum integer length of the shortest side.
Solution:
Let the shortest side be x.
Then longest side length = 2x
and third side length = x + 2
Perimeter = x + 2x + x + 2 = 4x +2
Given, perimeter > 166
∴ 4x + 2 > 166
∴ 4x > 164
∴ x > 41
∴ Minimum integer length of shortest side is 42 cm.

## Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Probability Miscellaneous Exercise 7

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Miscellaneous Exercise 7 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Miscellaneous Exercise 7

Question 1.
From a group of 2 men (M1, M2) and three women (W1, W2, W3), two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E?
Solution:
Let S be the sample space of the given event.
∴ S = {(M1, M2), (M1, W1), (M1, W2), (M1, W3), (M2, W1), (M2, W2), (M2, W3), (W1, W2) (W1, W3), (W2, W3)}
Let E be the event that one man and one woman are selected.
∴ E = {(M1, W1), (M1, W2), (M1, W3), (M2, W1), (M2, W2), (M2, W3)}
Here, the order is not important in which 2 persons are selected e.g. (M1, M2) is the same as (M2, M1)

Question 2.
Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that the three selected consist of 1 girl and 2 boys?
Solution:

Let G1, G2, G3 denote events for selecting a girl,
and B1, B2, B3 denote events for selecting a boy from 1st, 2nd and 3rd groups respectively.
Then P(G1) = $$\frac{3}{4}$$, P(G2) = $$\frac{2}{4}$$, P(G3) = $$\frac{1}{4}$$
P(B1) = $$\frac{1}{4}$$, P(B2) = $$\frac{2}{4}$$, P(B3) = $$\frac{3}{4}$$
Where G1, G2, G3, B1, B2 and B3 are mutually exclusive events.
Let E be the event that 1 girl and 2 boys are selected
∴ E = (G1 ∩ B2 ∩ B3) ∪ (B1 ∩ G2 ∩ B3) ∪ (B1 ∩ B2 ∩ G3)
∴ P(E) = P(G1 ∩ B2 ∩ B3) + P(B1 ∩ G2 ∩ B3) + P(B1 ∩ B2 ∩ G3)

Question 3.
A room has 3 sockets for lamps. From a collection of 10 light bulbs, 6 are defective. A person selects 3 at random and puts them in every socket. What is the probability that the room, will be lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in 10C3 ways.
∴ n(S) = 10C3
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in 6C3 ways.
∴ n(A’) = 6C3
∴ P(A’) = $$\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}$$
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 4.
There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Solution:
There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in
5C3 = $$\frac{5 \times 4 \times 3}{1 \times 2 \times 3}$$ = 10 ways.
∴ n(S) = 10
Let A be the event that 2 balls are red and 1 ball is black
2 red balls can be drawn out of 2 red balls in 2C2 = 1 way
and 1 black ball can be drawn out of 3 black balls in 3C1 = 3 ways.
∴ n(A) = 2C2 × 3C1 = 1 × 3 = 3
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{10}$$
Let B be the event that 1 ball is red and 2 balls are black
1 red ball out of 2 red balls can be drawn in 2C1 = 2 ways
and 2 black balls out of 3 black balls can be drawn in 3C2 = $$\frac{3 \times 2}{1 \times 2}$$ = 3 ways.
∴ n(B) = 2C1 × 3C2 = 2 × 3 = 6
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{10}$$
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B) = P(A) + P(B)
= $$\frac{3}{10}+\frac{6}{10}$$
= $$\frac{9}{10}$$

Question 5.
A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is even?
Solution:
Two tickets can be drawn out of 25 tickets in 25C2 = $$\frac{25 \times 24}{1 \times 2}$$ = 300 ways.
∴ n(S) = 300
Let A be the event that product of two numbers is even.
This is possible if both numbers are even, or one number is even and other is odd.
As there are 13 odd numbers and 12 even numbers from 1 to 25.
∴ n(A) = 12C2 + 12C1 × 13C1
= $$\frac{12 \times 11}{1 \times 2}$$ + 12 × 13
= 66 + 156
= 222
∴ Required probability = P(A)
= $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}$$
= $$\frac{222}{300}$$
= $$\frac{37}{50}$$

Question 6.
A, B and C are mutually exclusive and exhaustive events associated with the random experiment. Find P(A), given that
P(B) = $$\frac{3}{2}$$ P(A) and P(C) = $$\frac{1}{2}$$ P(B)
Solution:
P(B) = $$\frac{3}{2}$$ P(A) and P(C) = $$\frac{1}{2}$$ P(B)
Since A, B, C are mutually exclusive and exhaustive events,

Question 7.
An urn contains four tickets marked with numbers 112, 121, 122, 222, and one ticket is drawn at random. Let Ai (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A1, A2, and A3.
Solution:
One ticket can be drawn out of 4 tickets in 4C1 = 4 ways.
∴ n(S) = 4
According to the given information,
Let A1 be the event that 1st digit of the number of tickets is 1
A2 be the event that the 2nd digit of the number of tickets is 1
A3 be the event that the 3rd digit of the number of tickets is 1
∴ A1 = {112, 121, 122}, A2 = {112}, A3 = {121}

∴ A1, A2, A3 are not pairwise independent
For mutual independence of events A1, A2, A3
We require to have
P(A1 ∩ A2 ∩ A3) = P(A1) P(A2) P(A3)
and P(A1) P(A2) = P(A1 ∩ A2),
P(A2) P(A3) = P(A2 ∩ A3),
P(A1) P(A3) = P(A1 ∩ A3)
∴ From (iii),
A1, A2, A3 are not mutually independent.

Question 8.
The odds against a certain event are 5 : 2 and the odds in favour of another independent event are 6 : 5. Find the chance that at least one of the events will happen.
Solution:
Let A and B be two independent events.
Odds against A are 5 : 2
∴ the probability of occurrence of event A is given by
P(A) = $$\frac{2}{5+2}=\frac{2}{7}$$
Odds in favour of B are 6 : 5
∴ the probability of occurrence of event B is given by
P(B) = $$\frac{6}{6+5}=\frac{6}{11}$$
∴ P(at least one event will happen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B) ……[∵ A and B are independent events]

Question 9.
The odds against a husband who is 55 years old living till he is 75 is 8 : 5 and it is 4 : 3 against his wife who is now 48, living till she is 68. Find the probability that
(i) the couple will be alive 20 years hence
(ii) at least one of them will be alive 20 years hence.
Solution:
Let A be the event that husband would be alive after 20 years.
Odds against A are 8 : 5
∴ the probability of occurrence of event A is given by
P(A) = $$\frac{5}{8+5}=\frac{5}{13}$$
∴ P(A’) = 1 – P(A)
= 1 – $$\frac{5}{13}$$
= $$\frac{8}{13}$$
Let B be the event that wife would be alive after 20 years.
Odds against B are 4 : 3
∴ the probability of occurrence of event B is given by
P(B) = $$\frac{3}{4+3}=\frac{3}{7}$$
∴ P(B’) = 1 – P(B)
= 1 – $$\frac{3}{7}$$
= $$\frac{4}{7}$$
Since A and B are independent events
∴ A’ and B’ are also independent events
(i) Let X be the event that both will be alive after 20 years.
∴ P(X) = (A ∩ B)
∴ P(X) = P(A) . P(B)
= $$\frac{5}{13} \times \frac{3}{7}$$
= $$\frac{15}{91}$$

(ii) Let Y be the event that at least one will be alive after 20 years.
∴ P(Y) = P(at least one would be alive)
= 1 – P(both would not be alive)
= 1 – P(A’ ∩ B’)
= 1 – P(A’). P(B’)
= 1 – $$\frac{8}{13} \times \frac{4}{7}$$
= 1 – $$\frac{32}{91}$$
= $$\frac{59}{91}$$

Question 10.
Two throws are made, the first with 3 dice and the second with 2 dice. The faces of each die are marked with the number 1 to 6. What is the probability that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8?
Solution:
When 3 dice are thrown, then the sample space S1 has 6 × 6 × 6 = 216 sample points.
∴ n(S1) = 216
Let A be the event that the sum of the numbers is not less than 15.
∴ A = {(3, 6, 6), (4, 5, 6), (4, 6, 5), (4, 6, 6), (5, 4, 6), (5, 5, 5), (5, 5, 6), (5, 6, 4), (5, 6, 5), (5, 6, 6), (6, 3, 6), (6, 4, 5), (6, 4, 6), (6, 5, 4), (6, 5, 5), (6, 5, 6), (6, 6, 3), (6, 6, 4), (6, 6, 5), (6, 6, 6)}
∴ n(A) = 20
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}\left(\mathrm{S}_{1}\right)}=\frac{20}{216}=\frac{5}{54}$$
When 2 dice are thrown, the sample space S2 has 6 × 6 = 36 sample points.
∴ n(S2) = 36
Let B be the event that sum of numbers is not less than 8.
∴ B = {(2, 6), (3, 5), (3,6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}\left(\mathrm{S}_{2}\right)}=\frac{15}{36}=\frac{5}{12}$$
A ∩ B = Event that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8.
∴ A and B are independent events
∴ P(A ∩ B) = P(A) . P(B)
= $$\frac{5}{54} \times \frac{5}{12}$$
= $$\frac{25}{648}$$

Question 11.
Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy getting is 0.28. Find the probability that a student chosen at random will get first class.
Solution:
Let A be the event that student chosen is a boy
B be the event that student chosen is a girl
C be the event that student gets first class
∴ P(A) = $$\frac{2}{3}$$, P(B) = $$\frac{1}{3}$$
Probability of student getting first class, given that student is boy
Probability of student getting first class given that student is a girl, is
P(C/A) = 0.28 = $$\frac{28}{100}$$
and P(C/B) = 0.25 = $$\frac{25}{100}$$
∴ Required probability = P((A ∩ C) ∪ (B ∩ C))
Since A ∩ C and B ∩ C are mutually exclusive events
∴ Required probability = P(A ∩ C) + P(B ∩ C)
= P(A) . P(C/A) + P(B) . P(C/B)

Question 12.
A number of two digits is formed using the digits 1, 2, 3,……, 9. What is the probability that the number so chosen is even and less than 60?
Solution:
The number of two digits can be formed from the given 9 digits in 9 × 9 = 81 different ways.
∴ n(S) = 81
Let A be the event that the number is even and less than 60.
Since the number is even, the unit place of two digits can be filled in 4P1 = 4 different ways by any one of the digits 2, 4, 6, 8.
Also the number is less than 60, so tenth place can be filled in 5P1 = 5 different ways by any one of the digits 1, 2, 3, 4, 5.
∴ n(A) = 4 × 5 = 20
∴ Required probability = P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{81}$$

Question 13.
A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls each are made without replacement. Find the probability that the first drawing will give 3 white balls and the second drawing will give 3 red balls.
Solution:
Total number of balls = 8 + 5 = 13.
3 balls can be drawn out of 13 balls in 13C3 ways.
∴ n(S) = 13C3
Let A be the event that all 3 balls drawn are white.
3 white balls can be drawn out of 5 white balls in 5C3 ways.
∴ n(A) = 5C3
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{{ }^{5} C_{3}}{{ }^{13} C_{3}}=\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143}$$
After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls.
Let B be the event that the second draw of 3 balls are red.
∴ Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by
P(B/A) = $$\frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}=\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{15}$$
∴ Required probability = P(A ∩ B)
= P(A) . P(B/A)
= $$\frac{5}{143} \times \frac{7}{15}$$
= $$\frac{7}{429}$$

Question 14.
The odds against student X solving a business statistics problem are 8 : 6 and the odds in favour of student Y solving the same problem are 14 : 16
(i) What is the chance that the problem will be solved, if they try independently?
(ii) What is the probability that neither solves the problem?
Solution:
(i) Let A be the event that X solves the problem B be the event that Y solves the problem.
Since the odds against student X solving the problem are 8 : 6
∴ Probability of occurrence of event A is given by
P(A) = $$\frac{6}{8+6}=\frac{6}{14}$$
and P(A’) = 1 – P(A)
= 1 – $$\frac{6}{14}$$
= $$\frac{8}{14}$$
Also, the odds in favour of student Y solving the problem are 14 : 16
∴ Probability of occurrence of event B is given by
P(B) = $$\frac{14}{14+16}=\frac{14}{30}$$ and
P(B’) = 1 – P(B)
= 1 – $$\frac{14}{30}$$
= $$\frac{16}{30}$$
Now A and B are independent events.
∴ A’ and B’ are independent events.
∴ A’ ∩ B’ = Event that neither solves the problem
= P(A’ ∩ B’)
= P(A’) . P(B’)
= $$\frac{8}{14} \times \frac{16}{30}$$
= $$\frac{32}{105}$$
A ∪ B = the event that the problem is solved
∴ P(problem will be solved) = P(A ∪ B)
= 1 – P(A ∪ B)’
= 1 – P(A’ ∩ B’)
= 1 – $$\frac{32}{105}$$
= 1 – $$\frac{73}{105}$$

(ii) P (neither solves the problem) = P(A’ ∩ B’)
= P(A’) P(B’)
= $$\frac{8}{14} \times \frac{16}{30}$$
= $$\frac{32}{105}$$

## Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.4

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.4 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.4

Question 1.
Two dice are thrown simultaneously, if at least one of the dice shows a number 5, what is the probability that sum of the numbers on two dice is 9?
Solution:
When two dice are thrown simultaneously, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let A be the event that at least one die shows number 5.
∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ n(A) = 11
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{11}{36}$$
Let B be the event that sum of the numbers on two dice is 9.
∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)}
Also, A ∩ B = {(4, 5), (5, 4)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = $$\frac{n(A \cap B)}{n(S)}=\frac{2}{36}$$
∴ Probability of sum of numbers on two dice is 9, given that one dice shows number 5, is given by
P(B/A) = $$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{2}{36}}{\frac{11}{36}}=\frac{2}{11}$$

Question 2.
A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Solution:
When two dice are thrown simultaneously, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let A be the event that sum of the numbers is an even number.
∴ A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
∴ n(A) = 18
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{18}{36}$$
Let B be the event that sum of outcomes is a perfect square.
∴ B = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
Also, A n B= {(1, 3), (2, 2), (3, 1)}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}$$
∴ Probability of sum of the numbers is a perfect square, given that sum of numbers is an even number, is given by
P(B/A) = $$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{3}{36}}{\frac{18}{36}}=\frac{3}{18}=\frac{1}{6}$$

Question 3.
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Solution:
Two tickets can be drawn from 11 tickets with replacement in 11 × 11 = 121 ways.
∴ n(S) = 121
Let A be the event that the sum of two numbers is even.
The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn.
There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11.
∴ n(A) = 6 × 6 + 5 × 5
= 36 + 25
= 61
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{61}{121}$$
Let B be the event that the numbers tickets drawn are odd
∴ n(B) = 6 × 6 = 36
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}$$
Since 6 odd numbers are common between A and B.
∴ n(A ∩ B) = 6 × 6 = 36
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}$$
∴ Probability of both the numbers are odd, given that sum is even, is given by
P(B/A) = $$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{36}{121}}{\frac{61}{121}}=\frac{36}{61}$$

Question 4.
A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club card is drawn.
B: an ace card is drawn.
Determine whether events A and B are independent or not.
Solution:
One card can be drawn out of 52 cards in 52C1 ways.
∴ n(S) = 52C1
Let A be the event that a club card is drawn.
1 club card out of 13 club cards can be drawn in 13C1 ways.
∴ n(A) = 13C1
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}$$
Let B be the event that an ace card is drawn.
An ace card out of 4 aces can be drawn in 4C1 ways.
∴ n(B) = 4C1
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}$$
Since 1 card is common between A and B
∴ n(A ∩ B) = 1C1
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{1}{52}$$ …….(i)
∴ P(A) × P(B) = $$\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}} \times \frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13 \times 4}{52 \times 52}=\frac{1}{52}$$ …….(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Question 5.
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that,
(i) problem is solved?
(ii) problem is not solved?
(iii) exactly two students solve the problem?
Solution:
Let A be the event that student A can solve the problem.
B be the event that student B can solve the problem.
C be the event that student C can solve problem.
∴ P(A) = $$\frac{1}{3}$$, P(B) = $$\frac{1}{4}$$, P(C) = $$\frac{1}{5}$$
∴ P(A’) = 1 – P(A) = 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$
P(B’) = 1 – P(B) = 1 – $$\frac{1}{4}$$ = $$\frac{5}{4}$$
P(C’) = 1 – P(C) = 1 – $$\frac{1}{5}$$ = $$\frac{4}{5}$$
Since A, B, C are independent events
∴ A’, B’, C’ are also independent events
(i) Let X be the event that problem is solved.
Problem can be solved if at least one of the three students solves the problem.
P(X) = P(at least one student solves the problem)
= 1 – P(no student solved problem)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – $$\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}$$
= 1 – $$\frac{2}{5}$$
= $$\frac{3}{5}$$

(ii) Let Y be the event that problem is not solved
∴ P(Y) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
= $$\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}$$
= $$\frac{2}{5}$$

(iii) Let Z be the event that exactly two students solve the problem.
∴ P(Z) = P(A ∩ B ∩ C’) ∪ P(A ∩ B’ ∩ C) ∪ P(A’ ∩ B ∩ C)
= P(A) . P(B) . P(C’) + P(A) . P(B’) . P(C) + P(A’) . P(B) . P(C)
= $$\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right)$$
= $$\frac{4}{60}+\frac{3}{60}+\frac{2}{60}$$
= $$\frac{3}{20}$$

Question 6.
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Solution:
Let A be the event that man will be alive at 60.
∴ P(A) = 0.83
Let B be the event that a woman will be alive at 55.
∴ P(B) = 0.97
A ∩ B = Event that both will be alive.
Also, A and B are independent events
∴ P(both man and his wife will be alive) = P(A ∩ B)
= P(A) . P(B)
= 0.83 × 0.97
= 0.8051

Question 7.
In an examination, 30% of the students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subjects I and subject II. A student is selected at random, what is the probability that the student
(i) has failed in the subject I, if it is known that he is failed in subject II?
(ii) has failed in at least one subject?
(iii) has failed in exactly one subject?
Solution:
Let A be the event that the student failed in Subject I
B be the event that the student failed in Subject II
Then P(A) = 30% = $$\frac{30}{100}$$
P(B) = 20% = $$\frac{20}{100}$$
and P(A ∩ B) = 10% = $$\frac{10}{100}$$
(i) P (student failed in Subject I, given that he has failed in Subject II) = P(A/B)
$$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\left(\frac{10}{100}\right)}{\left(\frac{20}{100}\right)}=\frac{10}{20}=\frac{1}{2}$$

(ii) P(student failed in at least one subject) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= $$\frac{30}{100}+\frac{20}{100}-\frac{10}{100}$$
= 0.40

(iii) P(student failed in exactly one subject) = P(A) + P(B) – 2P(A ∩ B)
= $$\frac{30}{100}+\frac{20}{100}-2\left(\frac{10}{100}\right)$$
= 0.30

Question 8.
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Solution:
A be the event that first gun hits the target
B be the event that second gun hits the target
C be the event that third gun hits the target
P(A) = 0.5, P(B) = 0.6, P(C) = 0.8
∴ P(A’) = 1 – P(A) = 1 – 0.5 = 0.5
∴ P(B’) = 1 – P(B) = 1 – 0.6 = 0.4
∴ P(C’) = 1 – P(C) = 1 – 0.8 = 0.2
Now A, B, C are independent events
∴ A’, B’, C are also independent events.
∴ P (at least one hit is registered)
= 1 – P(no hit is registered)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – (0.5) (0.4) (0.2)
= 1 – 0.04
= 0.96

Question 9.
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that
(i) first is white and second is black?
(ii) one is white and the other is black?
Solution:
Total number of balls = 10 + 15 = 25
Let S be an event that two balls are drawn at random without replacement in succession
∴ n(S) = 25C1 × 24C1 = 25 × 24
(i) Let A be the event that the first ball is white and the second is black.
First white ball can be drawn from 10 white balls in 10C1 ways
and second black ball can be drawn from 15 black balls in 15C1 ways.
∴ n(A) = 10C1 × 15C1
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{10} \mathrm{C}_{1} \times{ }^{15} \mathrm{C}_{1}}{25 \times 24}=\frac{10 \times 15}{25 \times 24}=\frac{1}{4}$$

Question 10.
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, what is the probability that at least one ball is black?
Solution:
Total number of balls in the urn = 4 + 5 + 6 = 15
Two balls can be drawn without replacement in 15C2 = $$\frac{15 \times 14}{1 \times 2}$$ = 105 ways
∴ n(S) = 105
Let A be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in 4C1 = 4 ways
and 1 non-black ball can be drawn out of remaining 11 non-black balls in 11C1 = 11 ways
∴ 1 black and 1 non black ball can be drawn in 4 × 11 = 44 ways
Also, 2 black balls can be drawn from 4 black balls in 4C2 = $$\frac{4 \times 3}{1 \times 2}$$ = 6 ways
∴ n(A) = 44 + 6 = 50
∴ Required probability = P(A) = $$\frac{n(A)}{n(S)}=\frac{50}{105}$$ = $$\frac{10}{21}$$

Alternate Solution:
Total number of balls = 15
Required probability = 1 – P(neither of two balls is black)
Balls are drawn without replacement
Probability of first non-black ball drawn = $$\frac{11}{15}$$
Probability of second non-black ball drawn = $$\frac{10}{14}$$
Probability of neither of two balls is black = $$\frac{11}{15} \times \frac{10}{14}=\frac{11}{21}$$
Required probability = 1 – $$\frac{11}{21}$$ = $$\frac{10}{21}$$

Question 11.
Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Solution:
Total number of balls in the urn = 5 + 3 + 7 = 15
Out of these 12 are non-blue balls.
Two balls can be drawn from 15 balls without replacement in 15C2
= $$\frac{15 \times 14}{1 \times 2}$$
= 105 ways.
∴ n(S) = 105
Let A be the event that at least one ball is blue,
i.e., 1 blue and other non-blue or both are blue.
∴ n(A) = 3C1 × 12C1 + 3C2
= 3 × 12 + 3
= 36 + 3
= 39
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{39}{105}=\frac{13}{35}$$

Alternate solution:
Total number of balls in the urn = 15
Required probability = 1 – P(neither of two balls is blue)
Balls are drawn one by one without replacement.
Probability of first non-blue ball drawn = $$\frac{12}{15}$$
Probability of second non-blue ball drawn = $$\frac{11}{14}$$
Probability of neither of two ball is blue = $$\frac{12}{15} \times \frac{11}{14}=\frac{22}{35}$$
∴ Required probability = 1 – $$\frac{22}{35}$$ = $$\frac{13}{35}$$

Question 12.
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball ¡s drawn from each bag, what is the Probability that two balls are of the same colour?
Solution:
Let A be the event that a blue ball is drawn from each bag.
Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is
P(A) = $$\frac{4}{9} \times \frac{3}{10}$$
Let B be the event that a green ball is drawn from each bag.
Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is
P(B) = $$\frac{5}{9} \times \frac{7}{10}$$
Since both, the events are mutually exclusive and exhaustive events
∴ P(that both the balls are of the same colour) = P(both are of blue colour) or P(both are of green colour)
= P(A) + P(B)
= $$\frac{4}{9} \times \frac{3}{10}+\frac{5}{9} \times \frac{7}{10}$$
= $$\frac{12}{90}+\frac{35}{90}$$
= $$\frac{47}{90}$$

Question 13.
Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?
Solution:
Two cards are drawn from a pack of 52 cards with replacement.
∴ n(S) = 52 × 52
Let A be the event that two cards drawn are face cards.
First card from 12 face cards is drawn with replacement in 12C1 = 12 ways
and second face card is drawn from 12 face card in 12C1 = 12 ways after replacement.
∴ n(A) = 12 × 12
∴ P(that both the cards drawn are face cards) = P(A)
= $$\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}$$

## Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.3

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.3 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.3

Question 1.
Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let A be the event that sum of numbers on two dice is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{4}{36}$$
Let B be the event that number on second die is greater than or equal to number on first die.
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)}
∴ n(B) = 21
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{21}{36}$$
Now, A ∩ B = {(1, 4), (2, 3)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{36}$$
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{4}{36}+\frac{21}{36}-\frac{2}{36}$$
= $$\frac{23}{36}$$

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either red or face card?
Solution:
One card can be drawn from the pack of 52 cards in 52C1 = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that a red card is drawn Red card can be drawn in 26C1 = 26 ways
∴ n(A) = 26
∴ P(A) = $$\frac{26}{52}$$
Let B be the event that a black card is drawn
∴ Black card can be drawn in 26C1 = 26 ways.
∴ n(B) = 26
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}$$
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= $$\frac{26}{52}+\frac{26}{52}$$
= $$\frac{52}{52}$$
= 1

(ii) Let A be the event that a red card is drawn
∴ red card can be drawn in 26C1 = 26 ways
∴ n(A) = 26
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}$$
Let B be the event that a face card is drawn There are 12 face cards in the pack of 52 cards
∴ 1 face card can be drawn in 12C1 = 12 ways
∴ n(B) = 12
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}$$
There are 6 red face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{52}$$
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{26}{52}+\frac{12}{52}-\frac{6}{52}$$
= $$\frac{32}{52}$$
= $$\frac{8}{13}$$

Question 3.
Two cards are drawn from a pack of 52 cards. What is the probability that,
(i) both the cards are of the same colour?
(ii) both the cards are either black or queens?
Solution:
Two cards can be drawn from 52 cards in 52C2 ways.
∴ n(S) = 52C2
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that both cards are red.
∴ 2 red cards can be drawn in 26C2 ways.
∴ n(A) = 26C2
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}$$
Let B be the event that both cards are black.
∴ 2 black cards can be drawn in 26C2 ways
∴ n(B) = 26C2
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}$$
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= $$\frac{25}{102}+\frac{25}{102}$$
= $$\frac{25}{51}$$

(ii) Let A be the event that both cards are black.
∴ 2 black cards can be drawn in 26C2 ways.
∴ n(A) = 26C2
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}$$
Let B be the event that both cards are queens.
There are 4 queens in a pack of 52 cards
∴ 2 queen cards can be drawn in 4C2 ways.
∴ n(B) = 4C2
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}$$
There are two black queen cards.
∴ n(A ∩ B) = 2C2 = 1

Question 4.
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that
(i) number on the ticket is a perfect square or divisible by 4?
(ii) number on the ticket is a prime number or greater than 30?
Solution:
Out of the 50 tickets, a ticket can be drawn in 50C1 = 50 ways.
∴ n(S) = 50
(i) Let A be the event that the number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49}
∴ n(A) = 7
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{50}$$
Let B be the event that the number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
∴ n(B) = 12
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{50}$$
Now, A ∩ B = {4, 16, 36}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{50}$$
Required probability = P (A u B)
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{7}{50}+\frac{12}{50}-\frac{3}{50}$$
= $$\frac{8}{25}$$

(ii) Let A be the event that the number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
∴ n(A) = 15
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{15}{50}$$
Let B be the event that the number is greater than 30.
∴ B = {31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}
∴ n(B) = 20
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{20}{50}$$
Now, A ∩ B = {31, 37, 41, 43, 47}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}$$
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{15}{50}+\frac{20}{50}-\frac{5}{50}$$
= $$\frac{15+20-5}{50}$$
= $$\frac{3}{5}$$

Question 5.
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random
(i) passed at least one examination.
(ii) passed in exactly one examination.
(iii) failed in both examinations.
Solution:
Out of hundred students 1 student can be selected in 100C1 = 100 ways.
∴ n(S) = 100
Let A be the event that the student passed in the first examination.
Let B be the event that student passed in second examination.
∴ n(A) = 60, n(B) = 50 and n(A ∩ B) = 30

(i) P(student passed in at least one examination) = P(A ∪ B)
= P(A) + P(B) – P (A ∩ B)
= $$\frac{6}{10}+\frac{5}{10}-\frac{3}{10}$$
= $$\frac{4}{5}$$

(ii) P(student passed in exactly one examination) = P(A) + P(B) – 2.P(A ∩ B)
= $$\frac{6}{10}+\frac{5}{10}-2\left(\frac{3}{10}\right)$$
= $$\frac{1}{2}$$

(iii) P(student failed in both examinations) = P(A’ ∩ B’)
= P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – $$\frac{4}{5}$$
= $$\frac{1}{5}$$

Question 6.
If P(A) = $$\frac{1}{4}$$, P(B) = $$\frac{2}{5}$$ and P(A ∪ B) = $$\frac{1}{2}$$. Find the values of the following probabilities.
(i) P(A ∩ B)
(ii) P(A ∩ B’)
(iii) P(A’ ∩ B)
(iv) P(A’ ∪ B’)
(v) P(A’ ∩ B’)
Solution:
Here, P(A) = $$\frac{1}{4}$$, P(B) = $$\frac{2}{5}$$ and P(A ∪ B) = $$\frac{1}{2}$$
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= $$\frac{1}{4}+\frac{2}{5}-\frac{1}{2}$$
= $$\frac{3}{20}$$

(ii) P(A’ ∩ B’) = P(A) – P(A ∩ B)
= $$\frac{1}{4}-\frac{3}{20}$$
= $$\frac{1}{10}$$

(iii) P(A’ ∩ B) = P(B) – P(A ∩ B)
= $$\frac{2}{5}-\frac{3}{20}$$
= $$\frac{1}{4}$$

(iv) P(A’ ∪ B’) = P(A ∩ B)’ …..[De Morgan’s law]
= 1 – P(A ∩ B)
= 1 – $$\frac{3}{20}$$
= $$\frac{17}{20}$$

(v) P(A’ ∩ B’) = P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – $$\frac{1}{2}$$
= $$\frac{1}{2}$$

Question 7.
A computer software company is bidding for computer programs A and B. The probability that the company will get software A is $$\frac{3}{5}$$, the probability that the company will get software B is $$\frac{1}{3}$$ and the probability that company will get both A and B is $$\frac{1}{8}$$. What is the probability that the company will get at least one software?
Solution:
Let A be the event that the company will get software A.
∴ P(A) = $$\frac{3}{5}$$
Let B be the event that company will get software B.
∴ P(B) = $$\frac{1}{3}$$
Also, P(A ∩ B) = $$\frac{1}{8}$$
∴ P(the company will get at least one software) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= $$\frac{3}{5}+\frac{1}{3}-\frac{1}{8}$$
= $$\frac{72+40-15}{120}$$
= $$\frac{97}{120}$$

Question 8.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Solution:
One card can be drawn from the pack of 52 cards in 52C1 = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let A be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in 13C1 ways
∴ n(A) = 13C1
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{52}=\frac{13}{52}$$
Let B be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in 4C1 ways
∴ n(B) = 4C1
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{52}=\frac{4}{52}$$
There is one queen card out of 4 which is also a heart card
∴ n(A ∩ B) = 1C1
∴ P(A ∩ B) = $$\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{52}=\frac{1}{52}$$
∴ P(card is a heart or a queen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= $$\frac{13}{52}+\frac{4}{52}-\frac{1}{52}$$
= $$\frac{13+4-1}{52}$$
= $$\frac{16}{52}$$
∴ P(A ∪ B) = $$\frac{4}{13}$$

Question 9.
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in 7C4
= $$\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}$$
= 35 ways.
∴ n(S) = 35
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in 3C3 ways while a girl can be selected in 4C1 ways.
∴ n(A) = 3C3 × 4C1 = 4
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{4}{35}$$
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in 4C3 ways and a boy can be selected in 3C1 ways.
∴ n(B) = 4C3 × 3C1 = 12
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{12}{35}$$
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
= P(A) + P(B)
= $$\frac{4}{35}+\frac{12}{35}$$
= $$\frac{16}{35}$$

## Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.2

Question 1.
A fair die is thrown two times. Find the chance that
(i) product of the numbers on the upper face is 12.
(ii) sum of the numbers on the upper face is 10.
(iii) sum of the numbers on the upper face is at least 10.
(iv) sum of the numbers on the upper face is 4.
(v) the first throw gives an odd number and the second throw gives a multiple of 3.
(vi) both the times die to show the same number (doublet).
Solution:
If a fair die is thrown twice, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
(i) Let A be the event that the product of the numbers on uppermost face is 12.
∴ A = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(A) = 4
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{4}{36}=\frac{1}{9}$$

(ii) Let B be the event that sum of the numbers on uppermost face is 10.
∴ B = {(4, 6), (5, 5), (6, 4)}
∴ n(B) = 3
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}$$

(iii) Let C be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
∴ C = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), ( 6, 6)}
∴ n(C) = 6
∴ P(C) = $$\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$$

(iv) Let D be the event that sum of the numbers on uppermost face is 4.
∴ D = {(1, 3), (2, 2), (3, 1)}
∴ n(D) = 3
∴ P(D) = $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}=\frac{1}{12}$$

(v) Let E be the event that 1st throw gives an odd number and 2nd throw gives multiple of 3.
∴ E = {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6)}
∴ n(E) = 6
∴ P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$$

(vi) Let F be the event that both times die shows same number.
∴ F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(F) = 6
∴ P(F) = $$\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$$

Question 2.
Two cards are drawn from a pack of 52 cards. Find the probability that
(i) both are black.
(ii) both are diamonds.
(iii) both are ace cards.
(iv) both are face cards.
(vi) both are from the same suit.
(vii) both are from the same denomination.
Solution:
Two cards can be drawn from a pack of 52 cards in 52C2 ways.
∴ n(S) = 52C2
(i) Let A be the event that both the cards drawn are black.
The pack of 52 cards contains 26 black cards.
∴ 2 cards can be drawn from them in 26C2 ways
∴ n(A) = 26C2
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

(ii) Let B be the event that both the cards drawn are diamond.
There are 13 diamond cards in a pack of 52 cards.
∴ 2 diamond cards can be drawn from 13 diamond cards in 13C2 ways
∴ n(B) = 13C2
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

(iii) Let C be the event that both the cards drawn are aces.
In a pack of 52 cards, there are 4 ace cards.
∴ 2 ace cards can be drawn from 4 ace cards in 4C2 ways
∴ n(C) = 4C2
∴ P(C) = $$\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

(iv) Let D be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
∴ 2 face cards can be drawn from 12 face cards in 12C2 ways.
∴ n(D) = 12C2
∴ P(D) = $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

(v) Let E be the event that out of the two cards drawn one is a spade and other is non-spade.
There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
∴ One spade card can be drawn from 13 spade cards in 13C1 ways and one non-spade card can be drawn from 39 non-spade cards in 39C1 ways.
∴ n(E) = 13C1 . 39C1
∴ P(E) = $$\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \cdot{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}$$

(vi) Let F be the event that both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
2 cards can be drawn from a suit in 13C2 ways.
A suit can be selected in 4 ways.
∴ n(F) = 13C2 × 4
∴ P(F) = $$\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

(vii) Let G be the event that both the cards drawn are of same denominations.
A pack of cards has 13 denominations and 4 different cards for each denomination
∴ n(G) = 13 × 4C2
∴ P(G) = $$\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}=\frac{13 \times{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}$$

Question 3.
Four cards are drawn from a pack of 52 cards. Find the probability that
(i) 3 are Kings and 1 is Jack.
(ii) all the cards are from different suits.
(iii) at least one heart.
(iv) all cards are club and one of them is a jack.
Solution:
4 cards can be drawn out of 52 cards in 52C4 ways.
∴ n(S) = 52C4
(i) Let A be the event that out of the four cards drawn, 3 are kings and 1 is a jack.
There are 4 kings and 4 jacks in a pack of 52 cards.
∴ 3 kings can be drawn from 4 kings in 4C3 ways.
Similarly, 1 jack can be drawn out of 4 jacks in 4C1 ways.
∴ Total number of ways in which 3 kings and 1 jack can be drawn is 4C3 × 4C1
∴ n(A) = 4C3 × 4C1
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}$$

(ii) Let B be the event that all the cards drawn are of different suits.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ A card can be drawn from each suit in 13C1 ways.
∴ 4 cards can be drawn from 4 different suits in 13C1 × 13C1 × 13C1 × 13C1 ways.
∴ n(B) = 13C1 × 13C1 × 13C1 × 13C1
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}$$

(iii) Let C be the event that out of the four cards drawn at least one is a heart.
∴ C’ is the event that all 4 cards drawn are non-heart cards.
In a pack of 52 cards, there are 39 non-heart cards.
∴ 4 non-heart cards can be drawn in 39C4 ways.
∴ n(C’) = 39C4
∴ P(C’) = $$\frac{\mathrm{n}\left(\mathrm{C}^{\eta}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}$$
∴ P(C) = 1 – P(C’) = 1 – $$\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}$$

(iv) Let D be the event that all the 4 cards drawn are clubs and one of them is a jack.
In a pack of 52 cards, there are 13 club cards having 1 jack card.
∴ 1 jack can be drawn in 1C1 way and the other 3 cards can be drawn from remaining 12 club cards in 12C3 ways.
∴ n(D) = 12C3 × 1C1
∴ P(D) = $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{3} \times{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}$$

Question 4.
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is 1/3. The probability of yellow is 1/5.
(i) What is the probability of blackball?
(ii) How many balls are green, black, and yellow?
Solution:
(i) The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow (Y)
∴ P(G) = $$\frac{1}{3}$$ and P(Y) = $$\frac{1}{5}$$
If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
∴ P(G) + P(B) + P(Y) = 1
∴ $$\frac{1}{3}$$ + P(B) + $$\frac{1}{5}$$ = 1
∴ P(B) + $$\frac{8}{15}$$ = 1
∴ P(B) = 1 – $$\frac{8}{15}$$ = $$\frac{7}{15}$$
∴ Probability of black ball is $$\frac{7}{15}$$

(ii) Total number of balls = 15 and
P(G) = $$\frac{1}{3}$$, P(B) = $$\frac{7}{15}$$, P(Y) = $$\frac{1}{5}$$
∴ number of green balls = $$\frac{1}{3}$$ × 15 = 5
number of black balls = $$\frac{7}{15}$$ × 15 = 7
and number of yellow balls = $$\frac{1}{5}$$ × 15 = 3.

Question 5.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that the
(i) number on the ticket is divisible by 6?
(ii) number on the ticket is a perfect square?
(iii) number on the ticket is prime?
(iv) number on the ticket is divisible by 3 and 5?
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in 75C1 = 75 ways.
∴ n(S) = 75
(i) Let A be the event that number on ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12}{75}=\frac{4}{25}$$

(ii) Let B be the event that number on ticket is a perfect square.
∴ B = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{8}{75}$$

(iii) Let C be the event that the number on the ticket is a prime number.
∴ C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(C) = 21
∴ P(C) = $$\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{21}{75}=\frac{7}{25}$$

(iv) Let D be the event that number on ticket is divisible by 3 and 5 i.e., divisible by L.C.M. of 3 and 5 i.e., 15
∴ D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{5}{75}=\frac{1}{15}$$

Question 6.
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains
(i) 3 boys and 2 girls
(ii) at least 3 boys.
Solution:
The group consists of 8 boys and 5 girls i.e., 8 + 5 = 13 persons.
A committee of 5 is to be formed from this group.
∴ 5 persons from 13 persons can be selected in 13C5 ways
∴ n(S) = 13C5
(i) Let A be the event that the committee contains 3 boys and 2 girls.
3 boys from 8 boys can be selected in 8C3 ways and 2 girls from 5 girls can be selected in 5C2 ways
∴ n(A) = 8C3 . 5C2
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}}{{ }^{13} \mathrm{C}_{5}}$$

(ii) Let B be the event that the committee contains at least 3 boys (i.e., 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
∴ n(B) = 8C3 . 5C2 + 8C4 . 5C1 + 8C5 . 5C0
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}+{ }^{8} \mathrm{C}_{4} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{5} \cdot{ }^{5} \mathrm{C}_{0}}{{ }^{13} \mathrm{C}_{5}}$$

Question 7.
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in a socket. What is the probability that the room is lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in 10C3 ways.
∴ n(S) = 10C3
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in 6C3 ways.
∴ n(A’) = 6C3
∴ P(A’) = $$\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}$$
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 8.
The letters of the word LOGARITHM are arranged at random. Find the probability that
(i) Vowels are always together.
(ii) Vowels are never together.
(iii) Exactly 4 letters between G and H
(iv) begins with O and ends with T
Solution:
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.
∴ n(S) = 9!
(i) Let A be the event that vowels are always together.
The word LOGARITHM consists of 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).
3 vowels can be arranged among themselves in = 3P3 = 3! ways.
Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7) can be arranged in 7P7 = 7! ways.
∴ n(A) = 3! × 7!
∴ P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3 ! \times 7 !}{9 !}$$

(ii) Let B be the event that vowels are never together.
Consider the following arrangement
_C_C_C_C_C_C_
6 consonants create 7 gaps.
∴ 3 vowels can be arranged in 7 gaps in 7P3 ways.
Also 6 consonants can be arranged among themselves in 6P6 = 6! ways.
∴ n(B) = 6! × 7P3
∴ P(B) = $$\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6 ! \times{ }^{7} \mathrm{P}_{3}}{9 !}$$

(iii) Let C be the event that exactly 4 letters are arranged between G and H.
Consider the following arrangement
1 2 3 4 5 6 7 8 9
∴ Out of 9 places, G and H can occupy any one of following 4 positions in 4 ways.
1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th
Now, G and H can be arranged among themselves in 2P2 = 2! =2 ways.
Also, the remaining 7 letters can be arranged in remaining 7 places in 7P7 = 7! ways.
∴ n(C) = 4 × 2 × 7! = 8 × 7! = 8!
∴ P(C) = $$\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{8 !}{9 !}=\frac{8 !}{9 \times 8 !}=\frac{1}{9}$$

(iv) Let D be the event that word begins with O and ends with T.
Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in 7P7 = 7! ways
∴ n(D) = 7! × 1 × 1 = 7!
∴ P(D) = $$\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{7 !}{9 !}$$

(v) Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in 7P7 = 7! ways
∴ n(E) = 3 × 6 × 7!
∴ P(E) = $$\frac{n(E)}{n(S)}=\frac{3 \times 6 \times 7 !}{9 !}$$

Question 9.
The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Solution:
The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T).
6 letters in which A repeats twice can be arranged among themselves in $$\frac{6 !}{2 !}$$ ways.
∴ n(S) = $$\frac{6 !}{2 !}$$
Let A be the event that vowels are always together.
3 vowels (A, A, I) can be arranged among themselves in $$\frac{3 !}{2 !}$$ ways.
Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in 4P4 = 4! ways.

## Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.1

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.1

Question 1.
State the sample space and n(S) for the following random experiments.
(i) A coin is tossed twice. If a second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.
Solution:
(i) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown.
Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
∴ S = {HH, TH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
∴ n(S) = 14

(ii) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
Let A be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
∴ A = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event B.
∴ B = {HTT, HTH, TTT, TTH}
The sample space S of the experiment is A ∪ B.
∴ S = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6, HTT, HTH, TTT, TTH}
∴ n(S) = 16

Question 2.
In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Solution:
The bag contains 3 balls out of which one is black (B), one is red (R) and the other one is green (G).
Two balls are drawn one after the other, with replacement, from the bag.
∴ the sample space S is given by
S = {BB, BR, BG, RB, RR, RG, GB, GR, GG}
∴ n(S) = 9

Question 3.
A coin and die are tossed. State sample space of following events.
(i) A: Getting a head and an even number.
(ii) B: Getting a prime number.
(iii) C: Getting a tail and perfect square.
Solution:
When a coin and a die are tossed the sample space S is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) A: getting a head and an even number
∴ A = {H2, H4, H6}

(ii) B: getting a prime number
∴ B = {H2, H3, H5, T2, T3, T5}

(iii) C: getting a tail and a perfect square.
∴ C = {T1, T4}

Question 4.
Find the total number of distinct possible outcomes n(S) for each of the following random experiments.
(i) From a box containing 25 lottery tickets and 3 tickets are drawn at random.
(ii) From a group of 4 boys and 3 girls, any two students are selected at random.
(iii) 5 balls are randomly placed into five cells, such that each cell will be occupied.
(iv) 6 students are arranged in a row for photographs.
Solution:
(i) Let S be the event that 3 tickets are drawn at random from 25 tickets
∴ 3 tickets can be selected in 25C3 ways
∴ n(S) = 25C3
= $$\frac{25 \times 24 \times 23}{3 \times 2 \times 1}$$
= 2300

(ii) There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in 7C2 ways.
∴ n(S) = 7C2
= $$\frac{7 \times 6}{2 \times 1}$$
= 21

(iii) 5 balls have to be placed in 5 cells in such a way that each cell is occupied.
∴ The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
∴ a total number of ways of filling 5 cells such that each cell is occupied = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(iv) Six students can be arranged in a row for a photograph in 6P6 = 6! ways.
∴ n(S) = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question 5.
Two dice are thrown. Write favourable outcomes for the following events.
(i) P: The sum of the numbers on two dice is divisible by 3 or 4.
(ii) Q: sum of the numbers on two dice is 7.
(iii) R: sum of the numbers on two dice is a prime number.
Also, check whether
(a) events P and Q are mutually exclusive and exhaustive.
(b) events Q and R are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, all possible outcomes are
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(i) P: sum of the numbers on two dice is divisible by 3 or 4.
∴ P = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(ii) Q: sum of the numbers on two dice is 7.
∴ Q = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(iii) R: sum of the numbers on two dice is a prime number.
∴ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
(a) P and Q are mutually exclusive events as P ∩ Q = φ and
P ∪ Q = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ P and Q are not exhaustive events as P ∪ Q ≠ S.

(b) Q ∩ R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ Q ∩ R ≠ φ
Also, Q ∪ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} ≠ S
∴ Q and R are neither mutually exclusive nor exhaustive events.

Question 6.
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits
(i) is not taken into account.
(ii) is taken into account.
Solution:
(i) If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in 52C1 = 52 ways.
∴ n(S) = 52

(ii) If consideration of suits is taken into account, then one card can be drawn from each suit in 13C1 × 4C1
= 13 × 4
= 52 ways.
∴ n(S) = 52

Question 7.
Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box II contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
∴ the sample space is
S = {(H, R11), (H, R12), (H, R13), (H, B11), (H, B12), (T, R21), (T, R22), (T, B21), (T, B22), (T, B23), (T, B24)}

Question 8.
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn
(i) with replacement
(ii) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8.
Two cards are to be drawn from this bag.
(i) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6,8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(ii) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

## Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Miscellaneous Exercise 6

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Miscellaneous Exercise 6

Question 1.
Find the value of r if 56Cr+6 : 54Pr-1 = 30800 : 1
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R
The number of words starting with A = 5!
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
The number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
Find the number of ways of distributing n balls in n cells. What will be the number of ways if each cell must be occupied?
Solution:
There are n balls and n cells
(i) Every ball can be put in any of the n cells.
Number of distributions = n × n × …… × n = (n)n
(ii) For filling the first cell, n balls are available.
The first cell is filled in n ways.
The second cell is filled in (n – 1) ways
The third cell is filled in (n – 2) ways and so on.
the nth cell is tilled in one way.
Required number = n(n – 1)(n – 2) …… 1 = n!

Question 5.
Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Solution:
Taking CST as the first station and Thane as 20th,
Let us name CST as A0 next station as A1 and so on, Thane is A20
From station A0, 20 different journeys are possible
From station A1, 20 different journeys are possible.
From station A20, 20 different journeys are possible.
Total number of different tickets of different journeys = 21 × 20 = 420

Question 6.
English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
Number of 3 Letter passwords = 11P3
= 11 × 10 × 9
= 990

Question 7.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= $$\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}$$
= $$\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}$$
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360
∴ 360 numbers that exceed one million can be formed with the digits 3, 2, 0, 4, 3, 2, 3.

Question 8.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:

Required number = 26C6 + 26C10

Question 9.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:

Required Number = 5 + 10 = 15

Question 10.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so. Solution:
Required number = 30C7 × 23C10 × 13C13

Question 11.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 27 = 128
This number includes one case when the student passes in all subjects.
∴ Required number = 128 – 1 = 127

Question 12.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:

Question 13.
Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Solution:
Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then A ∪ B ∪ C represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number = Total number of distributions – n(A ∪ B ∪ C) …..(i)
n(A ∪ B ∪ C) represent the number of undesirable cases
Total number of distributions = 3 × 3 × 3 × 3 × 3 = 35 = 243 …….(ii)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C) …..(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
∴ n(A) + n(B) + n(C) = 3 × 25 ……(iv)
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 3 × 15 ……(v)
∴ n(A ∩ B ∩ C) = 0 …….(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
n(A ∪ B ∪ C) = 3 × 25 – 3 × 15
= 96 – 3
= 93
Substitute n(A ∪ B ∪ C) and from (ii) to (i), we get
Required number = 243 – 93 = 150

Question 14.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 212
This number includes the case in which all 12 lamps are OFF.
∴ Required Number = 212 – 1 = 4095

Question 15.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
Let the quadratic equation be ax2 + bx + c = 0, a ≠ 0

∴ Required number = 3 × 4 × 4 = 48

Question 16.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
Let the telephone number be 45abcd

∴ Required number = 8P4 = 1680

Question 17.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 question.
Number of outcomes = 26
This number includes the case when the student solves NONE of the question.
Required number = 26 – 1
= 64 – 1
= 63

Question 18.
Find the number of ways of dividing 20 objects in three groups of sizes 8, 7 and 5.
Solution:
Select 8 objects out of 20 in 20C8 ways
Select 7 objects from the remaining 12 in 12C7 ways and 5 objects from the remaining 5 in 5C5 ways
Required number is = 20C8 × 12C7 × 5C5

Question 19.
There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number = 12C6 = 924

Question 20.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
Solution:
We need 2 lines from each set.
Required number = 4C2 × 5C2
= 6 × 10
= 60

## Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.7

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.7 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.7

Question 1.
Find n if nC8 = nC12
Solution:
nC8 = nC12
If nCx = nCy, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

Question 2.
Find n if 23C3n = 23C2n+3
Solution:
23C3n = 23C2n+3
If nCx = nCy, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

Question 3.
Find n if 21C6n = $${ }^{21} C_{n^{2}+5}$$
Solution:
21C6n = $${ }^{21} C_{n^{2}+5}$$
If nCx = nCy, then either x = y or x = n – y
∴ 6n = n2 + 5 or 6n = 21 – (n2 + 5)
∴ n2 – 6n + 5 = 0 or 6n = 21 – n2 – 5
∴ n2 – 6n + 5 = 0 or n2 + 6n – 16 = 0
If n2 – 6n + 5 = 0 then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then n2 + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n2 + 6n – 16 = 0 then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2

Question 4.
Find n if 2nCr-1 = 2nCr+1
Solution:
2nCr-1 = 2nCr+1
If nCx = nCy, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Question 5.
Find n if nCn-2 = 15
Solution:
nCn-2 = 15
nC2 = 15 …….[∵ nCr = nCn-r]
∴ $$\frac{n !}{(n-2) ! 2 !}$$ = 15
∴ $$\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) ! 2 \times 1}$$ = 15
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Comparing both sides, we get
∴ n = 6

Question 6.
Find x if nPr = x nCr
Solution:

Question 7.
Find r if 11C4 + 11C5 + 12C6 + 13C7 = 14Cr
Solution:

Question 8.
Find the value of $$\sum_{r=1}^{4}{ }^{21-r} C_{4}+{ }^{17} C_{5}$$
Solution:

Question 9.
Find the differences between the largest values in the following:
(i) 14Cr12Cr
Solution:
Greatest value of 14Cr
Here n = 14, which is even
Greatest value of nCr occurs at r = $$\frac{n}{2}$$ if n is even

∴ Difference between the greatest values of 14Cr and 12Cr = 3432 – 924 = 2508

(ii) 13Cr8Cr
Solution:
Greatest value of 13Cr
Here n = 13, which is odd
Greatest value of nCr occurs at r = $$\frac{\mathrm{n}-1}{2}$$ if n is odd

∴ Difference between the greatest values of 13Cr and 8Cr = 1716 – 70 = 1646

(iii) 15Cr11Cr
Solution:
Greatest value of 15Cr
Here n = 15, which is odd
Greatest value of nCr occurs at r = $$\frac{\mathrm{n}-1}{2}$$ if n is odd

Difference between the greatest values of 15Cr and 11Cr = 6435 – 462 = 5973

Question 10.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more join the party = 10 + 5 + 1 = 16

Question 11.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consists of at least 3 women.

∴ Number of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 12.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in the majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women
Consider the following table:

∴ Number of committees = 14112 + 14700 + 6720 + 1260 + 81 = 36873
∴ At least 5 women are there in 36873 committees.

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= 18C10 – 36873
= 18C8 – 36873
= 43758 – 36873
= 6885

Question 13.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 14.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicketkeeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected = 45045 + 6006 = 51051

Question 15.
Five students are selected from 11. How many ways can these students be selected if
(i) two specified students are selected?
(ii) two specified students are not selected?
Solution:
5 students are to be selected from 11 students
(i) When 2 specified students are included
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = 9C3
= $$\frac{9 !}{3 ! \times 6 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}$$
= 84
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

(ii) When 2 specified students are not included then 5 students can be selected from the remaining (11 – 2) = 9 students
∴ Number of ways of selecting 5 students from 9 students = 9C5
= $$\frac{9 !}{5 ! 4 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}$$
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not selected.

## Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.6

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.6

Question 1.
Find the value of
(i) 15C4
Solution:

(ii) 80C2
Solution:

(iii) 15C4 + 15C5
Solution:

(iv) 20C1619C16
Solution:

Question 2.
Find n if
(i) 6P2 = n 6C2
Solution:

(ii) 2nC3 : nC2 = 52 : 3
Solution:

(iii) nCn-3 = 84
Solution:
nCn-3 = 84
∴ $$\frac{n !}{(n-3) ![n-(n-3)] !}$$ = 84
∴ $$\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}$$ = 84
∴ n(n – 1) (n – 2) = 84 × 6
∴ n(n – 1) (n – 2) = 9 × 8 × 7
Comparing on both sides, we get
∴ n = 9

Question 3.
Find r if 14C2r : 10C2r-4 = 143 : 10
Solution:

∴ $$\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}$$
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if.
(i) nPr = 720 and nCn-r = 120
Solution:

(ii) nCr-1 : nCr : nCr+1 = 20 : 35 : 42
Solution:

Question 5.
If nPr = 1814400 and nCr = 45, find r.
Solution:

∴ r! = 40320
∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
∴ r! = 8!
∴ r = 8

Question 6.
If nCr-1 = 6435, nCr = 5005, nCr+1 = 3003, find rC5.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in 6C3 ways.
3 green balls can be selected from 5 green balls in 5C3 ways.
3 blue balls can be selected from 7 blue balls in 7C3 ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in 6C3 ways.
2 girls can be selected from 4 girls in 4C2 ways.
∴ Number of ways the team can be selected = 6C3 × 4C2
= $$\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}$$
= $$\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}$$
= 20 × 6
= 120
∴ The team of 3 boys and 2 girls can be selected in 120 ways.

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = nC2
Given 66 handshakes were exchanged.
∴ 66 = nC2
∴ 66 = $$\frac{n !}{2 !(n-2) !}$$
∴ 66 × 2 = $$\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}$$
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points = 20C2
= $$\frac{20 !}{2 ! 18 !}$$
= $$\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$$
= 190

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides. .
∴ Through ‘n’ points we can draw nC2 lines including sides.
∴ Number of diagonals in n sided polygon = nC2 – n (∴ n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = 20C2
= $$\frac{20 !}{2 ! 18 !}$$
= $$\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$$
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 10 points on a plane.
(i) No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear = 10C2
= $$\frac{10 !}{2 ! 8 !}$$
= $$\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}$$
= 5 × 9
= 45

(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
= 10C24C2 + 1
= 45 – $$\frac{4 !}{2 ! 2 !}$$ + 1
= 45 – $$\frac{4 \times 3 \times 2 !}{2 \times 2 !}$$ + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 12 points on the plane
(i) When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = 12C3
= $$\frac{12 !}{3 ! 9 !}$$
= $$\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}$$
= 220

(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points = 12C34C3
= 220 – $$\frac{4 !}{3 ! \times 1 !}$$
= 220 – $$\frac{4 \times 3 !}{3 !}$$
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
Out of 8 consonants, 4 can be selected in 8C4
= $$\frac{8 !}{4 ! 4 !}$$
= $$\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}$$
= 70 ways
From 3 vowels, 2 can be selected in 3C2
= $$\frac{3 !}{2 ! 1 !}$$
= $$\frac{3 \times 2 !}{2 !}$$
= 3 ways
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in 6P6 i.e., 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.