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Three squares of chess board are selected at random. Find the probability of getting 2 squares of one colour and other of a different colour.
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In a chess board, there are 64 squares of which 32 are white and 32 are black. Since 2 of one colour and 1 of other can be $2 \mathrm{~W}, 1 \mathrm{~B}$, or $1 \mathrm{~W}, 2 \mathrm{~B}$, the number of ways is
$\left({ }^{32} \mathrm{C}_2 \times{ }^{32} \mathrm{C}_1\right) \times 2$ and also, the number of ways of choosing any 3 boxes is ${ }^{64} \mathrm{C}_3$.
Hence, the required probability $=\frac{{ }^{32} \mathrm{C}_2 \times{ }^{32} \mathrm{C}_1 \times 2}{{ }^{64} \mathrm{C}_3}=\frac{16}{21}$
$\left({ }^{32} \mathrm{C}_2 \times{ }^{32} \mathrm{C}_1\right) \times 2$ and also, the number of ways of choosing any 3 boxes is ${ }^{64} \mathrm{C}_3$.
Hence, the required probability $=\frac{{ }^{32} \mathrm{C}_2 \times{ }^{32} \mathrm{C}_1 \times 2}{{ }^{64} \mathrm{C}_3}=\frac{16}{21}$
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