Three boxes contain balls as follows: • Box 1: 3 white, 1 black • Box 2: 2 white, 2 black • Box 3: 1 white, 3 black One ball is drawn at random from each box. What is the probability that exactly two whites and one black are drawn in total?

Introduction / Context:
We draw one ball independently from each of three differently composed boxes. The overall event “exactly two whites and one black” occurs via three mutually exclusive cases depending on which box contributes the black ball. Sum those case probabilities.

Given Data / Assumptions:

• Box 1: P(W) = 3/4, P(B) = 1/4.
• Box 2: P(W) = 2/4 = 1/2, P(B) = 1/2.
• Box 3: P(W) = 1/4, P(B) = 3/4.
• Independent draws across boxes.

Concept / Approach:

• Casework by which single box yields the black ball.
• Add the disjoint case probabilities (law of total probability).

Step-by-Step Solution:

Verification / Alternative check:
Complement check is longer here; the direct case-sum is clean and matches repeated simulation intuition (Box 3 is most likely to give black).

Why Other Options Are Wrong:

• 1/4 (= 8/32) and 3/16 (= 6/32) undercount; 1/32 counts only one case.
• “None of these” is false because 13/32 matches the exact calculation.

Common Pitfalls:

• Double-counting cases or forgetting that exactly one box must contribute black.

Final Answer:
13/32