Question

Three squares of a chess board are selected at random. The probability of getting 2 squares of one colour and other of different colour is
(a) $\frac{16}{21}$

(b) $\frac{8}{21}$

(c) $\frac{3}{32}$

(d) $\frac{3}{8}$

Answer 1

Number of ways of selecting 3 boxes from chess board = ⁶⁴C₃
[∴ chess has 8 × 8 = 64 square/boxes]
Numbers of ways of selecting 2 boxes of same colour = ³²C₂
Number of ways of selecting 1 box of other colour = ³²C₁
$\begin{array}{rcl}\mathrm{So},\mathrm{probability}& =& \frac{{}^{32}\mathrm{C}_2\times {}^{32}\mathrm{C}_1\times 2}{{}^{64}\mathrm{C}_3}\\ & =& \frac{16}{21}.\\ & & \end{array}$
Hence, the correct answer is option A.