Question

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 $2$ white and $1$ black balls will be drawn, is

• A. $\frac{13}{32}$
• B. $\frac{1}{4}$
• C. $\frac{1}{32}$
• D. $\frac{3}{16}$

Answer 1

Answer: (A)

$P (2$ white and $1$ black) $= P (W_1W_2B_3$ or $W_1B_2W_3$ or
$ B_1W_2W_3)$
$=P(W_1W_2B_3)+P(W_1B_2W_3)+P(B_1W_2W-3)$
$=P(W_1)P(W_2)P(B_3)+P(W_1)P(B_2)P(W_3)$
$ + P(B_1)P(W_2)P(W_3)$
=$\frac{3}{4}.\frac{2}{4}.\frac{3}{4}+\frac{3}{4}.\frac{2}{4}.\frac{1}{4}+\frac{1}{4}.\frac{2}{4}.\frac{1}{4}$
$=\frac{1}{32}.(9+3+1)=\frac{13}{32}.$