**Previous Exercise – Probability – Exercise 5.2**

#### Probability – Exercise 5.2

- There are 2 red and 2 yellow flowers in a basket. A child picks up at random three flowers. What is the probability of picking up both the yellow flowers?
- Shekar is one member of a group of 5 persons. If 3 out of these 5 persons is to be chosen for a committee, find the probability o shekar being in the committee.
- Three cards are drawn at random from a pack of 52 cards. What is the probability that all the three cards are kings?
- A committee of five persons is selected from 4 men and 3 women. What is the probability that the committee will have

(i) one man

(ii) two men

(iii) two women

(iv) at least two men

#### Probability – Exercise 5.2 – Solution

**There are 2 red and 2 yellow flowers in a basket. A child picks up at random three flowers. What is the probability of picking up both the yellow flowers?**

Solution:

There are 4 flowers. Out of those 3 flowers picked up by the child.

Then ^{4}C_{3} = ^{4!}/_{(4-3)!3!} = ^{3! x 4}/_{3!} = 4

n(S) =4

Let A be the event of picking the yellow flowers from the basket. There are only two yellow flowers in the basket (given)

n(A) = 2

P(A) = ^{n(A)}/_{n(S)} = ^{2}/_{4}

Therefore, the probability of picking yellow flowers = ^{2}/_{4}

**Shekar is one member of a group of 5 persons. If 3 out of these 5 persons is to be chosen for a committee, find the probability of shekar being in the committee.**

Solution:

Committee has to be created of 3 persons out of 5. Then, ^{5}C_{3} = 10

Let A be the event of choosing Shekar as a committee member. Then ^{5}C_{1} = 5

Therefore, the probability of shekar being in the committee = ^{n(A)}/_{n(S)} = ^{5}/_{10}

**Three cards are drawn at random from a pack of 52 cards. What is the probability that all the three cards are kings?**

Solution:

Three cards are drawn at random from a pack of 52 cards.

⸫ n(S) = ^{52}C_{3}

Let A be the event of getting all the three cards are kings.

Then, n(A) = ^{4}C_{3}

Therefore, the probability that all the three cards are kings = ^{n(A)}/_{n(S)} = ^{4C3}/_{52C3} = ^{4}/_{22100}

**A committee of five persons is selected from 4 men and 3 women. What is the probability that the committee will have**

**(i) one man**

**(ii) two men**

**(iii) two women**

**(iv) at least two men**

Solution:

A committee of five persons is selected from 4 men and 3 women.

n(S) = ^{7}C_{5} = 21

(i) Let A be the event of selecting one man out of 4 men

Then committee will have 1 man and 3 women . But we have 5 members in a group. Therefore, it is not possible to select one man.

n(A) = 0

Therefore, the probability that the committee will have if one man selected = ^{n(A)}/_{n(S)} = ^{0}/_{21} = 0

(ii) two men

Let B be the event of selecting two men out of 4 men.

Then committee will have 2 men and 3 women.

n(B) = ^{4}C_{2} x ^{3}C_{3} = 6 x 1 = 6

Therefore, the probability that the committee will have if two men selected, P(B) = ^{n(B)}/_{n(S)} = ^{6}/_{21} = ^{2}/_{7}

(iii) two women

Let C be the event of selecting two women out of 3 women.

Then committee will have 2 women and 3 men.

n(C) = ^{3}C_{2} x ^{4}C_{3} = 3 x 4 = 12

Therefore, the probability that the committee will have if two women selected , P(C) = ^{n(C)}/_{n(S)} = ^{12}/_{21} = ^{4}/_{7}

(iv) at least two men

Let D be the event of selecting at least two men that means committee may have more than two men. If committee has at least two men then comitte will have 2 men and 3 women ^{4}C_{2} x ^{3}C_{3} ; If committee has 3 men and 2 women then ^{4}C_{3} x ^{3}C_{2}, If committee has 4 men and 1 woman then, ^{4}C_{4} x ^{3}C_{1}

n(D) = ^{4}C_{2} x ^{3}C_{3} + ^{4}C_{3} x ^{3}C_{2 }+ ^{4}C_{4} x ^{3}C_{1} = 6 + 12 + 3 = 21

Therefore, the probability that the committee will have if comitte has atleast two men , P(D) = ^{n(D)}/_{n(S)} = ^{21}/_{21} = 1