A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let
a=P(X=3), b=P(X ≥ 3) and c=P(X ≥ 6 | X>3). Then (b+c)/a is equal to __________.

Solution

Maths sir · 5 steps

Write every event as a run of non-sixes; once X > 3 is given, only the extra failures count.

1. Find a. X = 3 means two non-sixes followed by a six. So a = (5/6)² × 1/6 = 25/216.
2. Find b. X ≥ 3 means the first two tosses are both non-sixes. What happens after that does not matter, so b = (5/6)² = 25/36.
3. Find c. In general P(X ≥ n) = (5/6)ⁿ⁻¹, and X > 3 is the same event as X ≥ 4. Given that, X ≥ 6 only needs tosses 4 and 5 to be non-sixes as well.

   c = P(X ≥ 6)/P(X ≥ 4) = (5/6)⁵/(5/6)³ = (5/6)² = 25/36

4. Combine. b + c = 50/36, so (b+c)/a = 50/36 × 216/25 = 2 × 6.
5. So. (b+c)/a = 12.

The common slip: 11. Taking c = (5/6)³, as if three more failures were needed after X > 3, gives 11. Only tosses 4 and 5 must fail, so c = (5/6)².

The idea to keep. For waiting till the first success, P(X ≥ n) = qⁿ⁻¹, and the process forgets the past: P(X ≥ m+n | X > m) = P(X ≥ n). Count only the extra failures you need.

Answer from the NTA final answer key. Solution by PrepWiser.