Wardrobe selection in the dark: Girdhari has 3 trousers (grey, blue, brown) and 4 shirts (1 grey, 3 white). He randomly picks one shirt–trouser pair without seeing the colors. What is the probability that neither the shirt nor the trouser is grey?

Introduction / Context:
This problem is a direct application of independent selection with simple fractions. We have a small wardrobe with categories (shirts, trousers) and want the probability that neither selected item is grey when picked at random in the dark.

Given Data / Assumptions:

• Trousers: 3 total (1 grey, 2 non-grey).
• Shirts: 4 total (1 grey, 3 non-grey).
• One shirt and one trouser are chosen independently and uniformly at random.

Concept / Approach:
For independence, multiply the probabilities of the required events in each category. The desired event is “non-grey shirt” AND “non-grey trouser.”

Step-by-Step Solution:

Verification / Alternative check:
List outcomes: trousers (G, B, Br), shirts (G, W, W, W). Favorable pairs exclude any with a G. Count = 2 * 3 = 6 favorable out of 3*4 = 12 total ⇒ 6/12 = 1/2.

Why Other Options Are Wrong:

• 1/12 or 1/6 severely undercount favorable outcomes.
• 1/4 double-filters incorrectly or ignores independence.

Common Pitfalls:
Treating the combined choice as dependent in the wrong way, or forgetting to multiply across categories for independent events.

Final Answer:
1/2