Question

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

• A. $\frac{2n}{{}^{2n}C_n}$
• B. $\frac{2^n}{{}^{2n}C_n}$
• C. $\frac{1}{{}^{2n}C_n}$
• D. None of these

Answer 1

The correct option is B

$\frac{2^n}{{}^{2n}C_n}$

Explanation for the correct option:

Step -1: Let S be the sample space and E is the event of selecting each pair consisting of one white and one red balls:

So, $n\left(E\right)=\left({}^{n}C_1{}^{n}C_1\right)\times \left({}^{n-1}C_1{}^{n-1}C_1\right).....\left({}^{2}C_1{}^{2}C_1\right)\times \left({}^{1}C_1{}^{1}C_1\right)$

$=\left(n\times n\right)\times \left(n-1\right)(n-1)....\left(2\times 2\right)\times \left(1\times 1\right)$

$=n^2\times {\left(n-1\right)}^2.....{\left(1\right)}^2$

$={(n!)}^2$

$n\left(S\right)={}^{2n}C_2\times {}^{2n-2}C_2\times {}^{2n-4}C_2\dots \dots {}^{4}C_2.{}^{2}C_2$

$=\frac{2n(2n-1)}{2}\times \frac{(2n-2)(2n-3)}{2}\times ....\frac{\left(4\times 3\right)}{2}\times \frac{\left(2\times 1\right)}{2}$

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

Step -2: The required probability

$\mathbf{Probability}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

$=\frac{{(n!)}^2}{{\displaystyle \frac{2n!}{2^n}}}$

$=\frac{2^n}{{\displaystyle \frac{2n!}{n!n!}}}$

$=\frac{2^{\mathrm{n}}}{{}^{2\mathrm{n}}\mathrm{C}_{\mathrm{n}}}$

Hence, option ‘B’ is correct.