A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Solution

Maths sir · 3 steps

Write the chance that the first 2 comes on throw n, then add up only the even n as a GP.

1. When 2 first appears. The first 2 comes on throw n only if the first n-1 throws are not 2, each with probability 5/6, and throw n is a 2, with probability 1/6. So P(n) = (5/6)ⁿ⁻¹1/6.
2. Keep even throws. Add P(2) + P(4) + P(6) + …. The first term is 5/6×1/6 = 5/36. Each next term has two more failures, so the common ratio is (5/6)² = 25/36.

   P(even) = (5/36)/(1-25/36) = 5/36×36/11 = 5/11

3. So. The probability that 2 appears on an even-numbered throw is 5/11, option (a).

The idea to keep. For 'repeat until success' questions, the first success on throw n has probability qⁿ⁻¹p. To get even or odd n only, sum a GP with ratio q².

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