There are two bags each containing 10different 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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$\frac{20 C_{10} + 2 (20 C_{11} + 20 C_{12})}{{(2^{10} - 1)}^{2}}$

$\frac{20 C_{10} + 2 (20 C_{11} + 20 C_{12}) - 110}{{(2^{10} - 1)}^{2}}$

$\frac{20 C_{10} + 2 (21 C_{12}) - 111}{{(2^{10} - 1)}^{2}}$

$\frac{20 C_{10} + 20 C_{11} + 2 (20 C_{12})}{{(2^{10} - 1)}^{2}}$

answer is D.

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Detailed Solution

No. of favourable ways

$\begin{array}{l} = \sum_{r = 1}^{10} 10 C_{r}^{2} + 2 \sum_{r = 1}^{9} 10 C_{r} 10 C_{r + 1} + 2 \sum_{r = 1}^{8} 10 C_{r} 10 C_{r + 2} \\ (difference0) (difference1) (difference2) \end{array}$

$(10 C_{10} - 1) + 2 (20 C_{11} - 10) + 2 (20 C_{8} - 10 C_{2}) = 20 C_{10} + 2 (20 C_{11} + 20 C_{12}) - 111$

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