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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
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Verified Answer
The correct answer is:
$13 / 12$
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
$$
\begin{array}{l}
=P(B W W)+P(W B W)+P(W W B) \\
=P(B) P(W) P(W)+P(W) P(B) P(W) \\
\quad+P(W) P(W) P(B) \\
=\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \\
=\frac{2+6+18}{64}=\frac{26}{64} \\
=\frac{13}{32}
\end{array}
$$
$$
(\mathrm{III})=1 W, 3 B .
$$
$\therefore$ Required probability
$$
\begin{array}{l}
=P(B W W)+P(W B W)+P(W W B) \\
=P(B) P(W) P(W)+P(W) P(B) P(W) \\
\quad+P(W) P(W) P(B) \\
=\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \\
=\frac{2+6+18}{64}=\frac{26}{64} \\
=\frac{13}{32}
\end{array}
$$
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