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. $1)\frac{13}{32}$
• B. $2)\frac{1}{4}$
• C. $3)\frac{1}{32}$
• D. none of these

Answer 1

Answer: (A)

$1)\frac{13}{32}$

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

Scenario 1: white ball from box 1, white ball from box 2, black ball from box 3

Scenario 2: white ball from box 1, black ball from box 2, white ball from box 3

Scenario 3: black ball from box 1, white ball from box 2, white ball from box 3

For box #1: P(white ball) = 3/4 = 0.75 and P(black ball) = 1/4 = 0.25

For box #2: P(white ball) = 2/4 = 0.5 and P(black ball)= 2/4 = 0.5

For box #3: P(white ball) = 1/4 = 0.25 and P(black ball) = 3/4 = 0.75

The probability of scenario 1 occurring is 0.75 * 0.5 * 0.75 = 0.28125

The probability of scenario 2 occurring is 0.75 * 0.5 * 0.25 = 0.09375

The probability of scenario 3 occurring is 0.25 * 0.5 * 0.25 = 0.03125

Thus, the probability that 2 white balls and 1 black ball will be drawn is P(scenario 1) + P(scenario 2) + P(scenario 3) = 0.28125 + 0.09375 + 0.03125 = 0.40625=13/32