Two fair dice and two fair coins are tossed simultaneously. What is the probability that both coins show heads and the sum of the two dice is a prime number?

Introduction / Context:
This is a product event: coin results are independent of dice results. We need both coins to be heads and the dice sum to be prime.

Given Data / Assumptions:

• Two coins → P(HH) = 1/4.
• Two dice → 36 equally likely ordered outcomes.
• Prime sums possible: 2, 3, 5, 7, 11.

Concept / Approach:
Compute the dice probability first, then multiply by 1/4 for HH.

Step-by-Step Solution:
Count prime-sum outcomes: sum 2 → 1 way; sum 3 → 2 ways; sum 5 → 4 ways; sum 7 → 6 ways; sum 11 → 2 ways.Total favorable on dice = 1 + 2 + 4 + 6 + 2 = 15.P(sum is prime) = 15/36 = 5/12.P(HH and prime sum) = (1/4) * (5/12) = 5/48.

Verification / Alternative check:
Check symmetry: sums 2 and 12 each have 1 way; 3 and 11 have 2; 5 and 9 have 4; 7 has 6. Prime pattern count 15 is standard.

Why Other Options Are Wrong:
5/72 and 13/144 reflect incorrect counts; 1/12 misses the coin factor; 7/144 is not supported by standard sum counts.

Common Pitfalls:
Forgetting that ordered dice pairs like (3,2) and (2,3) are distinct; or failing to multiply by the coin probability.

Final Answer:
5/48