Question

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

• A. $\frac{n!}{(2n!)}$
• B. $\frac{(n!{)}^2}{\left(2n\right)!}$
• C. $\frac{n!{.2}^2}{\left(2n\right)!}$
• D. $\frac{(n!{)}^2{2}^n}{\left(2n\right)!}$

Answer 1

Answer: (D)

$\frac{(n!{)}^2{2}^n}{\left(2n\right)!}$

Probability of the first pair having 1 white and 1 black = $n\ast n$
Probability of the second pair having 1 white and 1 black = $(n-1)\ast (n-1)$
Proceeding in this manner, the number of ways in which each pair has 1 white and 1 black = ${\{n\ast (n-1)\ast (n-2)...3\ast 2\ast 1\}}^2\ast 2^n$
(the $2^n$ comes because the order of white and black can be decided in $2!=2$ ways for each pair)
Therefore, required probability = $\frac{{\{n\ast (n-1)\ast (n-2)...3\ast 2\ast 1\}}^2\ast 2^n}{2n\ast (2n-1)\ast (2n-2)...3\ast 2\ast 1}=\frac{{(n!)}^2\ast 2^n}{\left(2n\right)!}$