Question

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

• A. $\frac{9}{23}$
• B. $\frac{7}{15}$
• C. $\frac{8}{21}$
• D. $\frac{9}{19}$

Answer 1

The correct option is D $\frac{9}{19}$

Number of ways in which $20$ persons can be divided into two equal groups $=\frac{20!}{10!10!2!}$
Number of ways in which $18$ people can be divided into groups of $10$ and $8=\frac{18!}{10!8!}$
Required probability $=\frac{\frac{18!}{10!8!}}{\frac{20!}{10!10!2!}}$
$\Rightarrow$ Required probability $=\frac{9}{19}$