Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.5 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 1.

If odds in favour of X solving a problem are 4 : 3 and odds against Y solving the same problem are 2 : 3. Find the probability of:

(i) X solving the problem

(ii) Y solving the problem

Solution:

(i) Odds in favour of X solving a problem are 4 : 3.

∴ The probability of X solving the problem is

P(X) = \(\frac{4}{4+3}=\frac{4}{7}\)

(ii) Odds against Y solving the problem are 2 : 3.

∴ The probability of Y solving the problem is

P(Y) = 1 – P(Y’)

= 1 – \(\frac{2}{2+3}\)

= 1 – \(\frac{2}{5}\)

= \(\frac{3}{5}\)

Question 2.

The odds against John solving a problem are 4 to 3 and the odds in favour of Rafi solving the same problem are 7 to 5. What is the chance that the problem is solved when both of them try it?

Solution:

The odds against John solving a problem are 4 to 3.

Let event P(A’) = P (John does not solve the problem)

= \(\frac{4}{4+3}\)

= \(\frac{4}{7}\)

So, the probability that John solves the problem

P(A) = 1 – P(A’) = 1 – \(\frac{4}{7}\) = \(\frac{3}{7}\)

Similarly, Let P(B) = P(Rafi solves the problem)

Since the odds in favour of Rafi solving the problem are 7 to 5,

P(B) = \(\frac{7}{7+5}\) = \(\frac{7}{12}\)

Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Since A, B are independent events,

P(A ∩ B) = P(A) . P(B)

∴ Required probability = P(A) + P(B) – P(A) . P(B)

Question 3.

The odds against student X solving a statistics problem are 8 : 6 and odds in favour of student Y solving the same problem are 14 : 16. Find the chance that

(i) the problem will be solved if they try it independently.

(ii) neither of them solves the problem.

Solution:

The odds against X solving a problem are 8 : 6.

Let P(X’) = P(X does not solve the problem) = \(\frac{8}{8+6}\) = \(\frac{8}{14}\)

So, the probability that X solves the problem

P(X) = 1 – P(X’) = 1 – \(\frac{8}{14}\) = \(\frac{6}{14}\)

Similarly, let P(Y) = P(Y solves the problem)

Since odds in favour of Y solving the problem are 14 : 16,

P(Y) = \(\frac{14}{14+16}=\frac{14}{30}\)

So, the probability that Y does not solve the problem

P(Y’) = 1 – P(Y)

= 1 – \(\frac{14}{30}\)

= \(\frac{16}{30}\)

(i) Required probability

P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)

Since X and Y are independent events,

P(X ∩ Y) = P(X) . P(Y)

∴ Required probability = P(X) + P(Y) – P(X) . P(Y)

= \(\frac{6}{14}+\frac{14}{30}-\frac{6}{14} \times \frac{14}{30}\)

= \(\frac{73}{105}\)

(ii) Required probability = P(X’ ∩ Y’)

Since X and Y are independent events, X’ and Y’ are also independent events.

∴ Required probability = P(X’) . P(Y’)

= \(\frac{8}{14} \times \frac{16}{30}\)

= \(\frac{32}{105}\)

Question 4.

The odds against a husband who is 60 years old, living till he is 85 are 7 : 5. The odds against his wife who is now 56, living till she is 81 are 5 : 3. Find the probability that

(i) at least one of them will be alive 25 years hence.

(ii) exactly one of them will be alive 25 years hence.

Solution:

The odds against her husband living till he is 85 are 7 : 5.

Let P(H’) = P(husband dies before he is 85) = \(\frac{7}{7+5}=\frac{7}{12}\)

So, the probability that the husband would be alive till age 85

P(H) = 1 – P(H’) = 1 – \(\frac{7}{12}\) = \(\frac{5}{12}\)

Similarly, P(W’) = P(Wife dies before she is 81)

Since the odds against wife will be alive till she is 81 are 5 : 3.

∴ P(W’) = \(\frac{5}{5+3}=\frac{5}{8}\)

So, the probability that the wife would be alive till age 81

P(W) = 1 – P(W’) = 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)

(i) Required probability

P(H ∪ W) = P(H) + P(W) – P(H ∩ W)

Since H and W are independent events,

P(H ∩ W) = P(H) . P(W)

∴ Required probability = P(H) + P(W) – P(H) . P(W)

= \(\frac{5}{12}+\frac{3}{8}-\frac{5}{12} \times \frac{3}{8}\)

= \(\frac{40+36-15}{96}\)

= \(\frac{61}{96}\)

(ii) Required probability = P(H ∩ W’) + P(H’ ∩ W)

Since H and W are independent events, H’ and W’ are also independent events.

∴ Required probability = P(H) . P(W’) + P(H’) . P(W)

= \(\frac{5}{12} \times \frac{5}{8}+\frac{7}{12} \times \frac{3}{8}\)

= \(\frac{25+21}{96}\)

= \(\frac{46}{96}\)

= \(\frac{23}{48}\)

Question 5.

There are three events A, B, and C, one of which must, and only one can happen. The odds against event A are 7 : 4 and odds against event B are 5 : 3. Find the odds against event C.

Solution:

Since odds against A are 7 : 4,

P(A) = \(\frac{4}{7+4}=\frac{4}{11}\)

Since odds against B are 5 : 3,

P(B) = \(\frac{3}{5+3}=\frac{3}{8}\)

Since only one of the events A, B and C can happen,

P(A) + P(B) + P(C) = 1

\(\frac{4}{11}\) + \(\frac{3}{8}\) + P(C) = 1

∴ P(C) = 1 – (\(\frac{4}{11}\) + \(\frac{3}{8}\))

= 1 – \(\left(\frac{32+33}{88}\right)\)

= \(\frac{23}{88}\)

∴ P(C’) = 1 – P(C)

= 1 – \(\frac{23}{88}\)

= \(\frac{65}{88}\)

∴ Odds against the event C are P(C’) : P(C)

= \(\frac{65}{88}\) : \(\frac{23}{88}\)

= 65 : 23

Question 6.

In a single toss of a fair die, what are the odds against the event that number 3 or 4 turns up?

Solution:

When a fair die is tossed, the sample space is

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

Let event A: 3 or 4 turns up.

∴ A = {3, 4}

∴ n(A) = 2

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{6}=\frac{1}{3}\)

P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

∴ Odds against the event A are P(A’) : P(A)

= \(\frac{2}{3}: \frac{1}{3}\)

= 2 : 1

Question 7.

The odds in favour of A winning a game of chess against B are 3 : 2. If three games are to be played, what are the odds in favour of A’s winning at least two games out of the three?

Solution:

Let event A: A wins the game and event B: B wins the game.

Since the odds in favour of A winning a game against B are 3 : 2,

the probability of occurrence of event A and B is given by

P(A) = \(\frac{3}{3+2}=\frac{3}{5}\) and P(B) = \(\frac{2}{3+2}=\frac{2}{5}\)

Let event E: A wins at least two games out of three games.

∴ P(E) = P(A) . P(A) . P(B) + P(A) . P(B) . P(A) + P(B) . P(A) . P(A) + P(A) . P(A) . P(A)

∴ Odds in favour of A’s winning at least two games out of three are P(E) : P(E’)

= \(\frac{81}{125}: \frac{44}{125}\)

= 81 : 44