Question

A group of $6$ boys and $6$ girls is randomly divided into two equal groups. The probability that each group contains $3$ boys and $3$ girls is

• A. $10/231$
• B. $5/231$
• C. $90/231$
• D. $100/231$

Answer 1

The correct option is D $100/231$

The number of ways of choosing 6 persons out of 12 for a group is ${}^{12}{C}_6$.
The number of ways in which this group can contain 3 boys and 3 girls is ${}^6{C}_3{.}^6{C}_3$.
Therefore required probability
$=\frac{{}^6{C}_3{.}^6{C}_3}{{}^{12}{C}_6}=\frac{100}{231}$.