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 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
- 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
- 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
- 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) = 6/21 = 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) = 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) = 12/21 = 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 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