Nitish has 3 trousers (black, blue, brown) and 4 shirts (1 black, 3 white). In the dark he randomly picks one shirt–trouser pair. What is the probability that neither the shirt nor the trouser is black?

Introduction / Context:
This is a simple independent choice problem: one trouser is chosen from 3, and one shirt from 4, with equal likelihood. We need the probability that both chosen items are non-black.

Given Data / Assumptions:

• Trousers: {black, blue, brown} → 2 non-black out of 3.
• Shirts: {black, white, white, white} → 3 non-black out of 4.
• Choices are independent (one trouser and one shirt).

Concept / Approach:
P(neither black) = P(non-black trouser) * P(non-black shirt) = (2/3) * (3/4).

Step-by-Step Solution:
(2/3) * (3/4) = 6/12 = 1/2.

Verification / Alternative check:
Total outfit combinations = 3*4 = 12. Non-black trouser (2 choices) with non-black shirt (3 choices) = 6 combinations → 6/12 = 1/2.

Why Other Options Are Wrong:
1/12 and 1/6 are much too small; 1/4 ignores the independence product.

Common Pitfalls:
Counting only white shirts as non-black but forgetting there are three of them.

Final Answer:
1/2