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Q.

If 4 squares are chosen at random on a chess board, the probability that they lie on a diagonal line is

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a

\frac{2 \sum_{n = 4}^{8}^{n} C_{4}}{^{64} C_{4}}

b

\frac{2 \sum_{n = 4}^{8}^{n} C_{4} +^{8} C_{4}}{^{64} C_{4}}

c

\frac{4 \sum_{n = 4}^{8}^{n} C_{4}}{^{64} C_{4}}

d

\frac{4 \sum_{n = 4}^{7}^{n} C_{4} + 2 (^{8} C_{4})}{^{64} C_{4}}

answer is D.

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Detailed Solution

Question Image

4 squares can be chosen as ${2.}^{8} C_{4} + 4 \sum_{n = 4}^{7}^{n} C_{4} = n (E)$

$P (E) = \frac{4 \sum_{n = 4}^{7}^{n} C_{4} + 2 (^{8} C_{4})}{^{64} C_{4}}$

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If 4 squares are chosen at random on a chess board, the probability that they lie on a diagonal line is