Three squares of a chessboard are selected at random. The probability of selecting two squares of one colour…
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
\(\frac{10}{17}\)
\(\frac{15}{19}\)
\(\frac{17}{23}\)
\(\frac{16}{21}\)
Solution
Total number of ways of selecting 3 square
\(=64_{\mathrm{C}_3}\)
Total number of ways of selecting 2 square of one colour and other square of different colour \(=(2\) white, 1 black \()\) or (1 white, 2 black \()\)
\(\begin{aligned}
& ={ }^{32} \mathrm{C}_2 \cdot{ }^{32} \mathrm{C}_1+{ }^{32} \mathrm{C}_1 \cdot{ }^{32} \mathrm{C}_2 \\
& =2 \cdot 32_{\mathrm{C}_2} \cdot 32_{\mathrm{C}_1}
\end{aligned}\)
\(\begin{aligned}
\text {Required Probability } & =\frac{2.32_{\mathrm{C}_2} \cdot 32_{\mathrm{C}_1}}{64_{\mathrm{C}_3}} \\
& =\frac{2 \cdot \frac{32 \times 31}{2 \times 1} \times 32}{\frac{64 \times 63 \times 62}{3 \times 2 \times 1}}=\frac{16}{21}
\end{aligned}\)
Hence, option (d) is correct.