Question
There are two bags each containing 10 different books. A student draws out any number of books (at least one) from first bag as well as from the second bag. The probability that the difference between the number of books drawn from the two bags does not exceed two is
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Solution
Sol. [Key Concept: Two bags of 10 books each: count favorable selections.]
Number of favorable ways
$$= \sum_{r = 1}^{10}\mspace{2mu}\ ^{10}C_{r}^{2} + 2\sum_{r = 1}^{9}\mspace{2mu}\ ^{10}C_{r}^{10}C_{r + 1} + 2\sum_{r = 1}^{8}\mspace{2mu}\ ^{10}C_{r}^{10}C_{r + 2}$$
(difference 0) (difference 1) (difference 2)
$$\begin{matrix} & \ = \left( \ ^{20}C_{10} - 1 \right) + 2\left( \ ^{20}C_{11} - 10 \right) + 2\left( \ ^{20}C_{8} - \ ^{10}C_{2} \right) \\ & \ = \ ^{20}C_{10} + 2\ ^{20}C_{11} + 2\ ^{20}C_{12} - 111 \end{matrix}$$
no. of total ways $= \left( 2^{10} - 1 \right)^{2}$
$$P = \frac{\ ^{20}C_{10} + 2\left( \ ^{20}C_{11} + \ ^{20}C_{12} \right) - 111}{\left( 2^{10} - 1 \right)^{2}}$$
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