Probability – Exercise 5.3 – Class X

Previous Exercise – Probability – Exercise 5.2

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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

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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

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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 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

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3. 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

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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

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

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