Probability 3 Question 15
15. 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) $\frac{3}{16}$
(1998, 2M)
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Solution:
- $P(2$ white and 1 black $)=P\left(W _1 W _2 B _3\right.$ or $W _1 B _2 W _3$ or
$$ \begin{aligned} = & P\left(W _1 W _2 B _3\right)+P\left(W _1 B _2 W _3\right)+P\left(B _1 W _2 W _3\right) \\ = & P\left(W _1\right) P\left(W _2\right) P\left(B _3\right)+P\left(W _1\right) P\left(B _2\right) P\left(W _3\right) \\ & \quad+P\left(B _1\right) P\left(W _2\right) P\left(W _3\right) \\ =\frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4}+\frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4}= & \frac{1}{32}(9+3+1)=\frac{13}{32} \end{aligned} $$
$$ \left.B _1 W _2 W _3\right) $$
