Question

A committee of 2 people is to be chosen from 5 women and 5 men. What is the probability at least one man was chosen given at least one woman was chosen?

Answer 1

Answer:

5/7

Step-by-step explanation:

Number of women = 5

Number of men = 5

2 people is to be chosen to form a committee.

Let as assume

A : At least one man was chosen

B : At least one woman was chosen

A ∩ B : Exactly one man and one woman was chosen

P(B) = Exactly one woman was chosen + Two women was chosen

$P(B)=\dfrac{^5C_1\times ^5C_1+^5C_0\times ^5C_2}{^{10}C_2}\Rightarrow \frac{5\times 5+1\times 10}{45}=\frac{35}{45}=\frac{7}{9}$

$P(A\cap B)=\dfrac{^5C_1\times ^5C_1}{^{10}C_2}\Rightarrow \frac{5\times 5}{45}=\frac{25}{45}=\frac{5}{9}$

We need to find the probability at least one man was chosen given at least one woman was chosen.

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

$P(\frac{A}{B})=\dfrac{\frac{5}{9}}{\frac{7}{9}}$

$P(\frac{A}{B})=\frac{5}{7}$

Therefore, the required probability is 5/7.

Answer 2

Answer:

5/9

Step-by-step explanation:

At the beginning, there are 5 + 5 = 10 people to choose from. After one woman is chosen, that number goes down to 10 - 1  = 9 people. 5 men are still left to choose from, so there is a 5/9 chance at least one man will be selected as the other committee member.