Question

If one ball is drawn at random from each of the 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

The correct option is A $\frac{13}{32}$

Let's represent white balls with $W$ and black balls with $B$
box $1:3W1B$
box $2:2W2B$
box $3:1W3B$
Let $E$ be the event of drawing $2W$ and $1B$
then $P\left(E\right)=P\left(W\right)P\left(W\right)P\left(B\right)+P\left(W\right)P\left(B\right)P\left(W\right)+P\left(B\right)P\left(W\right)P\left(W\right)$
Here note that the order of the balls taken is written in synchronism with the order of the boxes $1,2$ and $3$.
$\Rightarrow P\left(E\right)=\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}+\frac{3}{4}\times \frac{2}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{26}{64}=\frac{13}{32}$