Question

Question 1 out of 80
A committee of 3 members is to be made out of 6 men and 5 women. What is the probability
that the committee has at least two women?
14/37
14/33
16/33
14/23​

Answer 1

Answer:

Probability = 14/33

Step-by-step explanation:

Given:

• A committee of 3 members is to be made out of 6 men and 5 women.

To Find:

• The probability that the committee has at least two women.

Solution:

First finding the total number of ways a committee of 3 can be formed.

Total number of persons = 6 men + 5 women = 11 persons

$\sf Total\:number\:of\:ways= \:^{11} C_3$

We know that,

$\sf ^nC_r=\dfrac{n!}{r!\times (n-r)!}$

Substitute the data,

$\sf Total\:number\:of\:ways= \dfrac{11!}{3!\times 8!}$

$\sf Total\:number\:of\:ways= \dfrac{11\times 10\times 9\times 8!}{3\times 2\times 8!}$

$\sf Total\:number\:of\:ways= 11\times 5\times 3$

= 165

Therefore the total number of ways a committee of 3 can be formed is 165.

Now number of ways that can a committee of 3 members can be formed consisting of at least 2 women is given by,

3 women or 2 women and one man

That is,

$\sf ^5C_3+\: ^5C_2\times \: ^6C_1$

Solving we get,

= 10 + 10 × 6

= 10 + 60

= 70

Now probability is given by,

Probability = Total number of ways committee consisting of atleast 2 women can be formed/Total number of ways a committee of 3 members can be formed.

Substitute the data,

Probability = 70/165

Probability = 14/33

Hence option 2 is correct.