Question

If one ball is drawn at random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, 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

$$\begin{gathered} \left(\mathrm{a}\right)\frac{13}{32} \\ \mathrm{Here},\mathrm{the}\mathrm{three}\mathrm{boxes}\mathrm{contain}3\mathrm{white}\mathrm{and}1\mathrm{black}\left(3W,1B\right),2\mathrm{white}\mathrm{and}2\mathrm{black}\left(2W,2B\right)\mathrm{and}1\mathrm{white}\mathrm{and}3blackballs\left(1W,3B\right),\mathrm{respectively}. \\ \mathrm{P}\left(2W,1B\right)=\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}+\frac{3}{4}\times \frac{2}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4} \\ =\frac{18}{64}+\frac{6}{64}+\frac{2}{64} \\ =\frac{26}{64} \\ =\frac{13}{32} \end{gathered}$$