Question

A bag contains $n$ red and $n$ white balls. Two balls at a time are drawn at random from the bag till all the ball are drawn. Find the probability that in each draw there is one white and one red ball.

• A. $\frac{\left(2^n\right)n!}{{}^{2n}{C}_n}$
• B. $\frac{4^n}{{}^{2n}{C}_n}$
• C. $\frac{2^n}{{}^{2n}{C}_n}$
• D. $\frac{\left(2^n\right)/n!}{{}^{2n}{C}_n}$

Answer 1

Answer: (C)

$\frac{2^n}{{}^{2n}{C}_n}$

Total number of balls $=2n(nwhite,nred)$

Total number of cases ${=}^{2n}{C}_2^{2n-2}{C}_2^{2n-4}{C}_2...{.}^2{C}_2$

$=\frac{\left(2n\right)!}{(2n-2)!2!}.\frac{(2n-2)!}{(2n-4)!2!}...\frac{2!}{2!0!}=\frac{\left(2n\right)!}{2^n}$

Now favourable ways.

In first draw white ball can be drawn in $n$ ways and red ball in $n$ ways. So both can be drawn in $n\times n=n^2$ ways.

Similarly when 2nd pair is drawn white $(n-1)$ ways and red in $(n-1)$ ways. So both in $(n-1)(n-1)={(n-1)}^2$ ways, so on.

Hence favourable cases $=n^2{(n-1)}^2...{.1}^2={[n(n-1)...1]}^2={(n!)}^2$

$\therefore$ The required probability $=\frac{{(n!)}^2}{\left(2n\right)!2^n}$ $=\frac{2^{n}n!n!}{\left(2n\right)!}=\frac{2^n}{{}^{2n}{C}_n}.$