  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
(2n)n!2nCn  B
4n2nCn  C
2n2nCn  D
(2n)/n!2nCn  Solution

The correct option is B $$\displaystyle \dfrac{2^{n}}{^{2n}C_{n}}$$Total number of balls $$=2n (n\ white, n\ red)$$Total number of cases $$\displaystyle =^{2n}C_{2} ^{2n-2}C_{2} ^{2n-4}C_{2}.... ^{2}C_{2}$$ $$\displaystyle =\dfrac{\left ( 2n \right )!}{\left ( 2n-2 \right )!2!}.\dfrac{\left ( 2n-2 \right )!}{\left ( 2n-4 \right )!2!}...\dfrac{2!}{2!0!}=\dfrac{\left ( 2n \right )!}{2^{n}}$$ 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 $$\displaystyle n\times n=n^{2}$$ ways.Similarly when 2nd pair is drawn white $$(n-1)$$ ways and red in $$(n-1)$$ ways. So both in $$\displaystyle \left ( n-1 \right )\left ( n-1 \right )=\left ( n-1 \right )^{2}$$ ways, so on.Hence favourable cases $$\displaystyle =n^{2}\left ( n-1 \right )^{2}....1^{2} =\left [ n\left ( n-1 \right )...1 \right ]^{2}=\left ( n! \right )^{2}$$$$\displaystyle \therefore$$ The required probability $$\displaystyle =\dfrac{\left ( n! \right )^{2}}{(2n)!2^{n}}$$ $$\displaystyle =\dfrac{2^{n}n!n!}{\left ( 2n \right )!}=\dfrac{2^{n}}{^{2n}C_{n}}.$$Maths

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