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.
• B.
• C.
• D. none

Answer 1

The correct option is A

The number of ways in which 20 people can be divided into two equals groups is
$n\left(s\right)=\frac{20!}{10!10!2!}$
The number of ways in which 18 people can be divided into groups of 10 and 8 is
$n\left(A\right)=\frac{18!}{10!8!}$
$\therefore P\left(E\right)=\frac{18!}{10!8!}\frac{10!10!2}{20!}=\frac{10\times 9\times 2}{20\times 19}=\frac{9}{19}$