Question

A committee consisting of 5 persons is to be chosen randomly from a group of 6men and 4 women what is the probability that exactly 2of the members of the committee are women

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A group consisting of 6 men and 4 women.

Total number of persons = 10

We know,

The number of ways of choosing 'r' objects from 'n' objects is

$\boxed{ \rm{ \: ^nC_r = \frac{n!}{r! \: (n - r)!} }}$

So,

Number of ways in which 5 persons can be selected from 10 persons is

$\rm \:  =  \:  \: ^{10}C_5$

$\rm \:  =  \:  \: \dfrac{10!}{5! \: (10 - 5)!}$

$\rm \:  =  \:  \: \dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times \: 5 \times 4 \times 3 \times 2 \times 1}$

$\rm \:  =  \:  \: 252$

Now,

We have to choose five persons consisting of exactly 2 women.

It means we have to choose 3 men out of 6 men and 2 women out of 4 women.

So,Number of ways to choose 2 women and 3 men out of 4 women and 6 men is

$\rm \:  =  \:  \: ^4C_2 \: \times \: ^6C_3$

$\rm \:  =  \:  \: \dfrac{4!}{2! \: (4 - 2)!} \times \dfrac{6!}{3! \: (6 - 3)!}$

$\rm \:  =  \:  \: \dfrac{4 \times 3 \times 2!}{2 \times 1 \: \times 2!} \times \dfrac{6 \times 5 \times 4 \times 3!}{3! \times \: 3 \times 2 \times 1}$

$\rm \:  =  \:  \: 6 \times 20$

$\rm \:  =  \:  \: 120$

So,

Required probability to choose 5 persons consisting of 2 women from 10 persons is

$\rm \:  =  \:  \: \dfrac{120}{252}$

$\rm \:  =  \:  \: \dfrac{30}{63}$

$\rm \:  =  \:  \: \dfrac{10}{21}$