ProbabilityHardBloom L3
Question
A bag consists of n white and n red balls. Pairs of balls are drawn without replacement until the bag is empty. Then the probability that each pair consists of one white and one red ball is
Options
A.$\frac{2^{n}n!}{\ ^{2n}C_{n}}$
B.$\frac{2^{n}}{\ ^{2n}C_{n}}$
C.$\frac{2^{n}}{(2n)!}$
D.$\frac{2^{n}}{(2n)!n!}$
Solution
Sol. [Key Concept: $n$ white $+ n$ red paired up: probability of the matching configuration.]
. Number of favorable ways $= n$ !
(distribute n red balls over n white balls)
no. of totals ways $= \frac{(2n)!}{(n!)(2!)^{n}}$
(divide $2n$ balls into ' $n$ ' equal groups)
$$\therefore\ p = \frac{n!}{\left( \frac{(2n)!}{(n!)(2!)^{n}} \right)} = \frac{2^{n}}{\ ^{2n}c_{n}}$$
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