A fair die is rolled three times and the sum of the three outcomes is 15. Given this condition, what is the probability that the first roll was a 4?

Introduction / Context:
This is a conditional probability on discrete uniform outcomes. We count the number of triples (x1, x2, x3) with xi ∈ {1,…,6} summing to 15, and among them count those with x1 = 4. The ratio gives the conditional probability.

Given Data / Assumptions:

• All ordered triples of die outcomes summing to 15 are equally likely.
• Die faces are 1 through 6.

Concept / Approach:
Use bounded-integer solution counting with inclusion–exclusion.

Step-by-Step Solution:
Let yi = xi − 1 so yi ∈ {0,…,5}. Then y1 + y2 + y3 = 12.Unrestricted nonnegative solutions: C(12 + 3 − 1, 3 − 1) = C(14, 2) = 91.Subtract cases where some yi ≥ 6: for any i, set zi = yi − 6 ≥ 0 → z-sum = 6 gives C(8, 2) = 28. There are 3 such i, subtract 3 × 28 = 84.Add back intersections where two yi ≥ 6: z-sum = 0 → C(2, 2) = 1, with C(3, 2) = 3 such intersections.Total valid triples = 91 − 84 + 3 = 10.Now impose x1 = 4 → x2 + x3 = 11. Pairs (x2, x3) in 1..6: (5,6) and (6,5) → 2 cases.Therefore conditional probability = 2 / 10 = 1/5.

Verification / Alternative check:
A quick list of all 10 triples confirms two begin with 4.

Why Other Options Are Wrong:
2/5 would require 4 cases out of 10; 1/6 assumes naive independence; “None of these” is false.

Common Pitfalls:
Forgetting that ordered triples are counted, or missing inclusion–exclusion adjustments.

Final Answer:
1/5