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 ball 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

$\frac{13}{32}$

Explanation for the correct answer:

There are three ways to draw 2 white balls and 1 black ball from the three boxes. They are:

Scenario 1: white ball from box 1, a white ball from box 2, and the black ball from box 3.

Scenario 2: white ball from box 1, a black ball from box 2, and the white ball from box 3.

Scenario 3: black ball from box 1, a white ball from box 2, and the white ball from box 3.

For box 1: $P$ (white ball)=$\frac{3}{4}$ and $P$ (black ball)=$\frac{1}{4}$

For box 2: $P$(white ball)=$\frac{2}{4}$ and $P$(black ball)=$\frac{2}{4}$

For box 3: $P$ (white ball)=$\frac{1}{4}$ and $P$ (black ball)=$\frac{3}{4}$

The probability of scenario one occurring is $\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}=\frac{18}{64}=\frac{9}{32}$

The probability of scenario 2 occurring is $\frac{3}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{6}{64}=\frac{3}{32}$

The probability of scenario 3 occurring is $\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{2}{64}=\frac{1}{32}$

Thus, the probability that 2 white balls and 1 black ball will be drawn is

$\mathrm{P}(\mathrm{scenario}1)+\mathrm{P}(\mathrm{scenario}2)+\mathrm{P}(\mathrm{scenario}3)=\frac{9}{32}+\frac{3}{32}+\frac{1}{32}$

$=\frac{13}{32}$

Hence, the correct answer is an option(A).