Question

A bag contains n white and n black balls. Pairs of balls are drawn without replacement until the bag is empty.The probability that each pair consists of one white and one black ball is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The required probability is
$\frac{n^2}{2n_{C_2}}\frac{(n-1{)}^2}{2n-2_{C_2}}\frac{(n-2{)}^2}{2n-4_{C_2}}....\frac{2^2}{4_{C_2}}\frac{1^2}{2_{C_2}}=\frac{(1\times 2\times 3\times 4\times ...\times (n-1)n{)}^2}{\frac{\left(2n\right)!}{2^n}}=\frac{2^n(n!{)}^2}{(2n!)}=\frac{2^n}{{}^{2n}{C}_n}$