Class 11

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 1.
What would be the simple interest on an amount of ₹ 9,600 at the rate of 6% per annum after 3 years?
Solution:
Given Principal P = ₹ 9600
Rate of interest R = 6% p.a.
Number of years = T = 3
Simple Interest I = \(\frac{\text { PRT }}{100}\)
= \(\frac{9600 \times 3 \times 6}{100}\)
= 96 × 18
= 1728
∴ Simple interest after 3 years would be ₹ 1728

Question 2.
What would be the simple interest at the rate of 9\(\frac{1}{2}\)% per annum on ₹ 6,000 for 2\(\frac{1}{2}\) years?
Solution:
Rate of interest per annum R = 9\(\frac{1}{2}\)% = \(\frac{19}{2}\)%
Principal P = ₹ 6000
Duration T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
∴ Simple Interest, I = \(\frac{\text { PRT }}{100}\)
= 6000 × \(\frac{19}{2} \times \frac{5}{2} \times \frac{1}{100}\)
= 15 × 19 × 5
= 1425
∴ Simple interest would be ₹ 1425.

Question 3.
What would be the simple interest on ₹ 8,400 in 9 months at the rate of 8.25 percent per annum?
Solution:
Principal P = ₹ 8400
Rate of interest R = 8.25%
Duration T = 9 months = \(\frac{3}{4}\) years

∴ Simple interest would be ₹ 519.75.

Question 4.
What would be the compound interest on ₹ 4200 for 18 months at 10% per annum compounded half yearly?
Solution:
Principal P = ₹ 4200
Rate of interest R = 10%
Duration T = 18 months = 1.5 years
compounding is done half yearly

= \(\frac{4200 \times 9261}{2000}\)
= 4862.025
I = A – P
= 4862.025 – 4200
= 662.025
∴ Compound interest would be ₹ 662.025.

Question 5.
Find compound interest on ₹ 10,000 for 2 years at 8% per annum compounded half yearly.
Solution:
Principal P = ₹ 10,000
Rate of interest R = 8% p.a. compounded half yearly
Duration T = 2 years

I = A – P
= 11648.58 – 10000
= 1698.58
∴ Compound interest is ₹ 1698.58.

Question 6.
In how many years ₹ 1,00,000 will become ₹ 1,33,100 at compound interest rate of 10% per annum?
Solution:
Principal P = ₹ 1,00,000
Amount A = ₹ 1,33,100
Rate of interest R = 10% p.a.

∴ ₹ 1,00,000 will become ₹ 1,33,100 after 3 years.

Question 7.
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many does it become double of itself at the same rate of simple interest?
Solution:
Given that, sum of money triples itself in 20 years
∴ P + I = 3P
∴ I = 2P
and T = 20 years
Now simple interest I = \(\frac{\text { PRT }}{100}\)
∴ 2P = \(\frac{\mathrm{P} \times \mathrm{R} \times 20}{100}\)
∴ R = 10
∴ Rate of interest = 10% per annum
The time period is to be calculated for the condition that the sum doubles itself i.e. for the condition
P + I = 2P
i.e. I = P
\(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = P
∴ \(\frac{10 \times T}{100}\) = 1
∴ T = 10
∴ The sum will become double of itself in 10 years.

Question 8.
A person borrows 10,000 for 2 year at 4% p.a. simple interest he immediately lends it to another person at 6.5% p.a. for 2 years. Find his total gain in the transaction.
Solution:
Person borrows money at 4% per annum and lends it at 6.5% per annum.
∴ His gain is (6.5 – 4) = 2.5% on ₹ 10000 for 2 years
i.e. gain = \(\frac{1000 \times 2.5 \times 2}{100}\)
= 100 × 5
= ₹ 500
∴ The person will gain ₹ 500 in this transaction.

Question 9.
A man deposits X 200 at the end of each year in recurring account at 5% compound interest. How much will it become at the end of 3 years?
Solution:
At end of 1st year, 2nd year and 3rd year ₹ 200 were deposited.
Rate of interest R = 5% p.a.
At end of 3 years, amount
= 200 + \(200\left[1+\frac{5}{100}\right]+200\left[1+\frac{5}{100}\right]^{2}\)
= 200 [1 + 1.05 + (1.05)2]
= 200 [2.05 + 1.1025]
= 200 [3.1525]
= 630.5
At end of 3 years, the account will have a balance of ₹ 630.5.

Question 10.
A man gets a simple interest of ₹ 2,000 on a certain principal at the rate of 5% p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate.
Solution:
Let Principal amount = P
Simple Interest I = ₹ 2000
Rate of interest R = 5% p.a.
Time duration T = 4 years
I = \(\frac{\text { PRT }}{100}\)
∴ 2000 = \(\frac{\mathrm{P} \times 5 \times 4}{100}\)
∴ P = 10000
Twice the principal was invested for compound interest with the same rate of interest for 2 years.
Here, P = 2 × 10,000 = ₹ 20,000
∴ Amount received,

I = A – P = 22050 – 20000 = 2050
The man will receive ₹ 2050 as compound interest.

Question 11.
The difference between simple interest and compound interest on a certain sum of money is ₹ 32 at 8% per annum for 2 years. Find the amount.
Solution:
Compound Interest = A – P = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-\mathrm{P}\)
Simple interest = \(\frac{\text { PRT }}{100}\)
Given R = 8%, T = 2 years and
compound interest – simple interest = ₹ 32

∴ The man will receive a compound interest of ₹ 5000.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 1.
Mr. Sarad purchased a laptop for ₹ 24,000 and sold it for ₹ 30,000. What was the profit percentage?
Solution:
Cost price (C.P.) = ₹ 24000
Selling price (S. P.) = ₹ 30,000
Profit = S.P. – C.P.
= 30,000 – 24,000
= 6,000

∴ Profit Percentage = 25%

Question 2.
Shraddha purchased a mobile phone and refrigerator for ₹ 18,000 and ₹ 15,000 respectively. She sold the refrigerator at a loss of 20% and the mobile at a profit of 20%. What is her overall profit or loss?
Solution:
C.P. of mobile phone = ₹ 18,000
Profit percentage on mobile phone = 20%
Selling price (S.P.) of mobile phone = 18,000 (1 + \(\frac{20}{100}\))
= 18,000 (1 + \(\frac{1}{5}\))
= 18,000 × \(\frac{6}{5}\)
= ₹ 21,600
C.P. of refrigerator = 15,000
Loss percentage on refrigerator = 20%
∴ Selling price (S.P.) = 15,000(1 – \(\frac{20}{100}\))
= 15,000(1 – \(\frac{1}{5}\))
= 15,000 × \(\frac{4}{5}\)
= ₹ 12,000
∴ Total gelling price for the transaction = 21,600 + 12,000 = ₹ 33,600
Total cost price (purchase price) for the transaction = 18,000 + 15,000 = ₹ 33,000
∴ Overall profit made by Shraddha = Total S.P. – Total C.P.
= 33,600 – 33,000
= ₹ 600
Thus, Shraddha made on overall profit of ₹ 600.

Question 3.
A vendor bought toffees at 6 for ₹ 10. How many for ₹ 10 must he sell to gain 20%?
Solution:
Vendor bought toffees at the rate of 6 for ₹ 10
∴ Cost price of one toffee = \(\frac{10}{6}\)
i.e. C.P. = \(\frac{10}{6}\) …….(i)
Let x be the number of toffees he must sell in ₹ 10 to gain 20%
i.e. S.P. = \(\frac{10}{x}\) …….(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }}\)
Using (i) and (ii) we have

∴ 30(6 – x) = 6x
∴ 180 – 30x = 6x
∴ 36x = 180
∴ x = 5
The vendor must sell 5 toffees for ₹ 10 in order to gain 20%.

Question 4.
The percentage profit earned by selling an article for ₹ 2,880 is equal to the percentage loss incurred by selling the same article for ₹ 1,920. At what price the article should be sold to earn a 25% profit?
Solution:
Let x be C.P. of the article
Let y % be both, the gain and loss made when article is sold at ₹ 2,880 and ₹ 1,920 respectively. Then
x + \(\frac{y}{100}\) x = 2880 ……(i)
x – \(\frac{y}{100}\) x = 1920 …..(ii)
Adding (i) and (ii), we get
2x = 4800
∴ x = 2400
i.e. C.P. of the article = ₹ 2400
Required profit percentage = 25%

∴ The article should be sold at ₹ 3000 to earn 25% profit.

Question 5.
A cloth merchant advertises for selling cloth at a 4% loss. By using a faulty meter scale, he is earning a profit of 20%. What is the actual length of the scale?
Solution:
Let the cost price of the cloth be ₹ ‘x’ per meter
He claims a loss of 4%
∴ Selling price of the cloth
S.P. = C.P.(1 – \(\frac{loss%}{100}\))
= x(1 – \(\frac{4}{100}\))
= 0.96x …..(i)
The actual cost price of the cloth is lower as the cloth is measured by a faulty meter scale.
Given that shopkeeper’s profit = 20%

∴ The actual cost price is 0.8 times the cost price as advertised.
In other words, the meter scale used for the fraud is 0.8 times the meter scale that should have been used.
∴ The length of the faulty meter scale used = 0.8 × 1 = 0.8 meter
∴ The actual length of the scale is 0.8 meters.

Question 6.
Sunil sells his bike worth ₹ 25,000 to Rohit at a profit of 20%. After 6 months Rohit sells the bike back to Sunil at a loss of 20%. Find the total profit percent of Sunil considering both the transactions.
Solution:
Sunil sells his bike to Rohit at 20% profit.
So S.P. of the bike for Sunil
= 25000 + \(\frac{20}{100}\) × 25000
= 25000 + 5000
= 30000
∴ Cost price of bike to Rohit = ₹ 30000
Rohit sells the bike back to Sunil at 20% loss
∴ S.P. of the bike for Rohit = 30000 – \(\frac{20}{100}\) × 30000
= 30000 – 6000
= 24000
∴ In second transaction Sunil pays 24000 to Rohit
In the first transaction, he had received 30000 from Rohit
∴ Sunil made a profit of ₹ (30000 – 24000) = ₹ 6000
Sunil earned this profit on the bike which costed him ₹ 25000
∴ Total profit % that Sunil makes = \(\frac{6000}{25000}\) × 100
= \(\frac{600}{25}\)
= 24
∴ Sunil makes 24% profit considering both the transactions.

Question 7.
By selling a book at ₹ 405 bookseller incurs a loss of 25%. Find the cost price of the book.
Solution:
S.P. = ₹ 405
Loss% = 25
S.P. when there is a loss is given by
S.P. = C.P. × \(\frac{\text { Loss } \%}{100}\)

∴ The cost price of the book is ₹ 540.

Question 8.
A cloth costs ₹ 675. If it is sold at a loss of 20%, what is its cost price as a percentage of its selling price?
Solution:
C.P. = ₹ 675
Loss% = 20%
∴ Loss made in selling = \(\frac{20}{100}\) × 675 = ₹ 135
S.P. = C.P. – Loss
= 675 – 135
= ₹ 540
Let C.P. be x % S.P.,
Then 675 = \(\frac{x}{100}\) × 540
∴ x = \(\frac{675 \times 100}{540}\) = 125
∴ Cost price is 125% of the selling price.

Question 9.
Ashwin buys an article for ₹ 500. He marks it for sale at 75% more than the cost price. He offers a 25% discount on the marked price to his customer. Calculate the actual percentage of profit made by Ashwin.
Solution:
C.P. = ₹ 500
Marked price = C.P. + \(\frac{75}{100}\) × C.P.
= \(\frac{75}{100}\) × 500
= 500 + 75 × 5
= 500 + 375
= 875
25% discount was given on marked price
∴ Discount = \(\frac{25}{100}\) × 875 = \(\frac{875}{4}\)
Selling price = marked price – discount

∴ Ashwin makes 31.25% profit.

Question 10.
The combined cost price of a refrigerator and a mixer is ₹ 12,400. If the refrigerator costs 600% more than the mixer, find the cost price of the mixer.
Solution:
Let ₹ x be the cost price of the mixer.
The cost price of the refrigerator = x + \(\frac{600}{100}\) x
= x + 6x
= 7x
Total cost price =12400 …..[Given]
i.e. x + 7x = 12400
i.e. 8x = 12400
∴ x = \(\frac{12400}{8}\) = 1550
∴ The cost price of mixer is ₹ 1550.

Question 11.
Find the single discount equivalent to the discount series of 5%, 7%, and 9%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{5}{100}\)) = 95
After 2nd discount the price = 95(1 – \(\frac{7}{100}\)) = \(\frac{95 \times 93}{100}\)
After 3rd discount the price = \(\frac{95 \times 93}{100}\left(1-\frac{9}{100}\right)\)
= \(\frac{95 \times 93 \times 91}{100 \times 100}\)
= \(\frac{803985}{10000}\)
= 80.3985 ~ 80.4
Selling price after 3 discounts is ₹ 80.4
Single equivalent discount = Marked price – Selling price
= 100 – 80.4
= ₹ 19.6
∴ Single equivalent discount is 19.6%.

Question 12.
The printed price of a shirt is ₹ 390. Lokesh pays ₹ 175.50 for it after getting two successive discounts. If the first discount is 10%, find the second discount.
Solution:
Marked price = ₹ 390
After the first discount of 10%, the price of the shirt

∴ x = 50
∴ Second discount is 50%

Question 13.
Amar, a manufacturer, gives a discount of 25% on the list price to his distributor Akbar, Akbar sells at a 10% discount on the list price to his customer Anthony. Anthony paid ₹ 540 for the article. What is the profit percentage of Akbar on his cost price?
Solution:
Let ₹ ‘x’ be the list price of the article.
Amar gives a discount of 25% on the list price.
∴ Selling price for Amar = \(x\left(1-\frac{25}{100}\right)\)
= \(x\left(1-\frac{1}{4}\right)\)
= ₹ \(\frac{3 x}{4}\)
Amar sells the article to Akbar
Cost price of article for Akbar = ₹ \(\frac{3 x}{4}\) ……(i)
Akbar sells the article to Anthony at 10% discount on list price
∴ Selling price for Akbar = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(1-\frac{1}{10}\right)\)
= ₹ \(\frac{9 x}{10}\) …..(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }} \times 100\)
Using (i) and (ii), we have the profit percentage for Akbar as,

∴ Akbar gets a profit of 20% on his cost price.

Question 14.
A man sells an article at a profit of 25%. If he had bought it at a 10% loss and sold it for ₹ 7 less, he would have gained 35%. Find the cost price of the article.
Solution:
Let ₹ ‘x’ be the C.P. of the article
∴ Article was sold at 25% profit
∴ S.P. of the article = \(x\left(1+\frac{25}{100}\right)\)
= \(x\left(1+\frac{1}{4}\right)\)
= 1.25x
If the article was bought at 10% loss
i.e., the new C.P. = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(\frac{9}{10}\right)\)
= 0.9x
and sold at ₹ 7 less
∴ New S.P. = 1.25x – 7
Then, the profit would have been 35%
Using profit percentage = \(\frac{\text { S.P.-C.P. }}{\text { C.P. }} \times 100\)

∴ Cost price of the article is ₹ 200

Question 15.
Mr. Mehta sold his two luxury cars at ₹ 39,10,000 each. On one he gains 15% but on the other, he loses 15%. How much does he gain or lose in the whole transaction?
Solution:
Let x, y be the C.P. of two cars.
S.P. of both the cars = 39,10,000 …..[Given]
∴ One car is sold at 15% loss
∴ S.P. of the first car = x – \(\frac{15}{100}\)x
∴ \(\frac{85}{100}\)x = 39,10,000
∴ x = \(\frac{39,10,000 \times 100}{85}\)
∴ x = 46,000 × 100
∴ x = 46,00,000
Other car is sold at 15% gain
∴ S.P. of second car = y + \(\frac{15}{100}\) y
∴ y + \(\frac{15}{100}\) y = 39,10,000
∴ \(\frac{115}{100}\)y = 39,10,000
∴ y = \(\frac{39,10,000 \times 100}{115}\)
∴ y = 34,000 × 100
∴ y = 34,00,000
x + y = Total C.P. of two cars
= 46,00,000 + 34,00,000
= 80,00,000
Total S.P. = 39,10,000 + 39,10,000 = 78,20,000
∴ S.P. < C.P.
∴ There is a loss of ₹ (80,00,000 – 78,20,000) = ₹ 1,80,000
∴ Loss % = \(\frac{1,80,000}{80,00,000} \times 100\)
= \(\frac{18}{8}\)
= 2.25
∴ Mr. Mehta bears a 2.25% loss in the whole transaction.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.1

Question 1.
Find 77% of 580 + 34% of 390.
Solution:

Question 2.
240 candidates appeared for an examination, of which 204 passed. What is the pass percentage?
Solution:
We find the pass percentage using the unitary method

∴ The pass percentage for the examination is 85%.

Question 3.
What percent of 8.4 kg are 168 grams?
Solution:
Let 168 gms be x% of 8.4 kg
i.e., let 168 gms be \(\frac{x}{100}\) of 8400 gms
∴ 168 = \(\frac{x}{100}\) × 8400
∴ x = \(\frac{168}{84}\) = 2
∴ 168 gms is 2% of 8.4 kg.

Question 4.
If the length of a rectangle is decreased by 20%, what should be the increase in the breadth of the rectangle so that the area remains the same?
Solution:
Let x and y represent the length and breadth of the rectangle respectively.
∴ The original area of the rectangle = xy
There is a 20% decrease in length.

Let k % be the required increase in breadth

Given that the new and old areas should be equal.

∴ 100 + k = 125
∴ k = 125 – 100 = 25
∴ Breadth should be increased by 25% so that the area remains same.

Question 5.
The price of rice increased by 20%, as a result, a person can have 5kg rice for ₹ 600. What was the initial price of rice per kg?
Solution:
A person can buy 5 kg of rice for ₹ 600 after the increase in price
∴ New price of rice = \(\frac{600}{5}\) = ₹ 120/kg …..(i)
Let ‘x’ be the initial price per kg of rice.
There is a 20% increase in the price of rice.
Thus the new price of the rice will be given as

∴ The initial price of rice is ₹ 100 per kg

Question 6.
What percent is 3% of 5%?
Solution:
Let 3% be x % of 5%.
Then \(\frac{3}{100}=\frac{x}{100} \times \frac{5}{100}\)
∴ x = \(\frac{3 \times 100}{5}\) = 60
∴ 3% is 60% of 5%.

Question 7.
After availing of two successive discounts of 20% each, Madhavi paid ₹ 64 for a book. If she would have got only one discount of 20%, how much additional amount would she have paid?
Solution:
Let the price of the book be ₹ x.
After the first 20% discount, the price of the book becomes

After another 20% discount, the price of the book becomes

This price = ₹ 64 …..[Given]
∴ \(\frac{16}{25}\)x = 64
∴ x = 4 × 25 = 100
Thus, Amount of the book after one discount = \(\frac{4}{5}\)(100) = 80 …..[from (i)]
∴ The additional amount that Madhavi would have paid = 80 – 64 = ₹ 16

Question 8.
The price of the table is 40% more than the price of a chair. By what percent price of a chair is less than the price of a table?
Solution:
Let ₹ x and ₹ y be the price of a table and chair respectively.
The price of the table is 40% more than the price of a chair
∴ \(\frac{x-y}{y}\) × 100 = 40

We need to find by how much percent is the price of a chair less than that of a table.

∴ The price of a chair is 28.57% less than the price of a table.

Question 9.
A batsman scored 92 runs which includes 4 boundaries 5 sixes. He scored other runs by running between the wickets. What percent of his total score did he make by running between the wickets?
Solution:
Batsman scores 4 fours (boundaries) and 5 sixes in 92 runs.
Number of runs scored by fours and sixes = 4 × 4 + 5 × 6 = 46
∴ 92 – 46 = 46
Let 46 be x% of 92.
Then 46 = \(\frac{x}{100}\) × 92
∴ x = \(\frac{46 \times 100}{92}=\frac{100}{2}\) = 50
∴ 50% of the total runs were scored by running between the wickets.

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Miscellaneous Exercise 8

Solve the following system of inequalities graphically.

Question 1.
x ≥ 3, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
2x + y ≥ 6, 3x + 4y < 12
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
x + y ≥ 4, 2x – y ≤ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
2x – y ≥1, x – 2y ≤ -1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
x + y ≤ 6, x + y ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 10.
3x + 4y ≤ 60, x +3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 14.
3x + 2y ≤ 150, x + 4y ≥ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.3

Find the graphical solution for the following system of linear inequations.

Question 1.
x – y ≤ 0, 2x – y ≥ -2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
2x + 3y ≥ 12, -x + y ≤ 3, x ≤ 4, y ≥ 3
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
3x + 2y ≤ 1800, 2x + 7y ≤ 1400
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
0 ≤ x ≤ 350, 0 ≤ y ≤ 150
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
\(\frac{x}{60}+\frac{y}{90}\) ≤ 1, \(\frac{x}{120}+\frac{y}{75}\) ≤ 1, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
3x + 2y ≤ 24, 3x + y ≥ 15, x ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

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.

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.

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 A_i (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’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

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)
= \(\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}\)

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)
= \(\frac{4}{35}+\frac{12}{35}\)
= \(\frac{16}{35}\)