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(D) Total number of ways of selecting $3$ squares from $64$ is $^{64}C_3$. Total number of ways of selecting $2$ squares of one colour and $1$ square

Vedclass · Accepted Answer

$\frac{16}{21}$

Vedclass · Answer

(D) Total number of ways of selecting $3$ squares from $64$ is $^{64}C_3$. Total number of ways of selecting $2$ squares of one colour and $1$ square of a different colour is $(2 	ext{ white, } 1 	ext{ black})$ or $(1 	ext{ white, } 2 	ext{ black})$. This is equal to $^{32}C_2 \cdot ^{32}C_1 + ^{32}C_1 \cdot ^{32}C_2 = 2 \cdot ^{32}C_2 \cdot ^{32}C_1$. Required Probability $= \frac{2 \cdot ^{32}C_2 \cdot ^{32}C_1}{^{64}C_3} = \frac{2 \cdot \frac{32 	imes 31}{2} 	imes 32}{\frac{64 	imes 63 	imes 62}{6}} = \frac{32 	imes 31 	imes 32 	imes 6}{64 	imes 63 	imes 62} = \frac{16}{21}$. Hence,option $D$ is correct.

Three squares of a chessboard are selected at random. The probability of selecting two squares of one colour and the other of a different colour is equal to

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