Question

There are only two women among $20$ persons taking part in a pleasure trip. The $20$ persons are divided into two groups, each group consisting of $10$ persons. Then the probability that the two women will be in the same group is:

• A. $9/19$
• B. $9/38$
• C. $9/35$
• D. none

Answer 1

The correct option is A $9/19$

$n\left(S\right)$ = number of ways in which 20 people can be divided into two equal groups

$⟹=\frac{20!}{10!\times 10!\times 2!}$

$n\left(A\right)$= 18 people can be divided into groups of 10 and 8 = $\frac{18!}{10!\times 8!}$

Hence, $P\left(\text{getting two women in same group}\right)=P\left(E\right)=\frac{\frac{18!}{10!8!}}{\frac{20!}{10!10!2!}}=\frac{18!10!10!2}{20!10!8!}=\frac{10!\times 2}{20\times 19\times 8!}=\frac{10\times 9\times 2}{380}$

Hence the required probability = $\frac{18}{38}=\frac{9}{19}$