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If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that two white and one black ball will be drawn is

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$\frac{13}{32}$

$\frac{1}{4}$

$\frac{1}{32}$

$\frac{3}{16}$

answer is A.

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Detailed Solution

$P$ ( $2$ white and $1$ black)

$= P (W_{1} W_{2} B_{3}$ or $W_{1} B_{2} W_{3}$ or $B_{1} W_{2} W_{3})$

$= P (W_{1}) P (W_{2}) P (B_{3}) + P (W_{1}) P (B_{2}) P (W_{3}) + P (B_{1}) P (W_{2}) P (W_{3})$

$= (\frac{3}{4}) (\frac{2}{4}) (\frac{3}{4}) + (\frac{3}{4}) (\frac{2}{4}) (\frac{1}{4}) + (\frac{1}{4}) (\frac{2}{4}) (\frac{1}{4})$
$= \frac{13}{32}$

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