Probability – Exercise 5.3 – Class X

Previous Exercise – Probability – Exercise 5.2

Probability – Exercise 5.2

1. 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?
2. 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.
3. Three cards are drawn at random from a pack of 52 cards. What is the probability that all the three cards are kings?
4. 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

1. 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 4C3 = 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

1. 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, 5C3 = 10

Let A be the event of choosing Shekar as a committee member. Then 5C1 = 5

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

1. 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) = 52C3

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

Then, n(A) = 4C3

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

1. 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) = 7C5 = 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) = 4C2 x 3C3 = 6 x 1 = 6

Therefore, the probability that the committee will have if two men selected, P(B) = n(B)/n(S) = 6/212/7

(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) = 3C2 x 4C3 = 3 x 4 = 12

Therefore, the probability that the committee will have if two women selected , P(C) = n(C)/n(S) = 12/214/7

(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  4C2 x 3C3 ; If committee  has 3 men and 2 women then 4C3 x 3C2, If committee has 4 men and 1 woman then, 4C4 x 3C1

n(D) = 4C2 x 3C3 + 4C3 x 3C2 + 4C4 x 3C1 = 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