A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x ≥ 5 | x > 2) is :

Solution

Maths sir · 3 steps

Turn both events into runs of non-sixes, then divide. The common failures cancel.

1. Rewrite the events. x ≥ 5 means the first 4 tosses are all non-sixes, so P(x ≥ 5) = (5/6)⁴. x > 2 means the first 2 tosses are non-sixes, so P(x > 2) = (5/6)².
2. Find the intersection. Every case with x ≥ 5 already has x > 2. So the event "x ≥ 5 and x > 2" is just x ≥ 5.

   P(x ≥ 5 | x > 2) = P(x ≥ 5)/P(x > 2) = (5/6)⁴/(5/6)² = (5/6)²

3. So. The conditional probability is (5/6)² = 25/36, which is option (d).

The idea to keep. For tossing until the first success, P(x > n) = qⁿ. Given x > m, the probability of x > n is just qⁿ⁻ᵐ, because the die has no memory of the earlier failures.

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