Q68 / 150 · MHT CET 2011 · Mathematics

Three boxes contain respectively 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, from each of the boxes one ball is drawn at random. The probability that 2 white and 1 black balls will be drawn, is

  • Correct answer
    A
    $13 / 12$
  • B
    $1 / 4$
  • C
    $1 / 32$
  • D
    $3 / 16$

Solution

Given, Box $(\mathrm{I})=3 W, 1 B ;$ Box $(\mathrm{II})=2 W, 2 B ;$ Box $$ (\mathrm{III})=1 W, 3 B . $$ $\therefore$ Required probability $$ \begin{array}{l} =P(B W W)+P(W B W)+P(W W B) \\ =P(B) P(W) P(W)+P(W) P(B) P(W) \\ \quad+P(W) P(W) P(B) \\ =\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \\ =\frac{2+6+18}{64}=\frac{26}{64} \\ =\frac{13}{32} \end{array} $$

Hence, the correct answer is option: 'A'.

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