CUET PG 2022 — Mathematics PYQ

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Question

CUET PG 2022 — Mathematics PYQ

CUET PG | Mathematics | 2022

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

Choose the correct answer:

Explanation

1. Identify Probabilities for Each Box

Let $W_i$ be the event of drawing a white ball and $B_i$ be the event of drawing a black ball from Box $i$ .

Box 1 (3W, 1B):

$P(W_1) = \frac{3}{4}, \quad P(B_1) = \frac{1}{4}$
Box 2 (2W, 2B):

$P(W_2) = \frac{2}{4} = \frac{1}{2}, \quad P(B_2) = \frac{2}{4} = \frac{1}{2}$
Box 3 (1W, 3B):

$P(W_3) = \frac{1}{4}, \quad P(B_3) = \frac{3}{4}$

2. Possible Combinations for (2 White, 1 Black)

There are three mutually exclusive ways to get exactly 2 white balls and 1 black ball:

Case 1: White from Box 1, White from Box 2, Black from Box 3 $(W_1, W_2, B_3)$
Case 2: White from Box 1, Black from Box 2, White from Box 3 $(W_1, B_2, W_3)$
Case 3: Black from Box 1, White from Box 2, White from Box 3 $(B_1, W_2, W_3)$

3. Calculation

The total probability $P(E)$ is the sum of these cases:

P(E) = [P(W_1) \cdot P(W_2) \cdot P(B_3)] + [P(W_1) \cdot P(B_2) \cdot P(W_3)] + [P(B_1) \cdot P(W_2) \cdot P(W_3)]

Substituting the values:

P(E) = \left( \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4} \right) + \left( \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} \right) + \left( \frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} \right)

P(E) = \frac{18}{64} + \frac{6}{64} + \frac{2}{64}

P(E) = \frac{18 + 6 + 2}{64}

P(E) = \frac{26}{64}

Final Answer:

By simplifying the fraction $\frac{26}{64}$ :

$P (E) = \frac{13}{32}$