First slide

Theorems of probability

Question

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

Moderate

A

$1 /^{2 n} C_{n}$
B

$2 n /^{2 n} C_{n}$
C

$2 n / n!$
D

$2 n / (2 n!)$

Solution

The required probability is
$\begin{array}{l} \frac{n^{2}}{^{2 n} C_{2}} \frac{(n - 1)^{2}}{2 n - 2} C_{2} \frac{(n - 2)^{2}}{^{2 n - 4} C_{2}} \dots \frac{2^{2}}{^{4} C_{2}} \frac{1^{2}}{^{2} C_{2}} \\ = \frac{(1 \times 2 \times 3 \times 4 \times \dots \times (n - 1) n)^{2}}{\frac{(2 n)!}{2^{n}}} = \frac{2^{n} (n!)^{2}}{(2 n)!} = \frac{2^{n}}{^{2 n} C_{n}} \end{array}$

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