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

This is a problem of geometric distribution. The probability of getting a 2 in a single throw of a fair die is 1/6.

Let's denote E as the event that 2 appears in an even number of throws.

We can break down the event E into two mutually exclusive events:

1. E1: 2 appears in the second throw. The probability of this event is (5/6) * (1/6) = 5/36.

2. E2: 2 does not appear in the first two throws, but appears in the fourth, sixth, eighth, etc. throw. The probability of this event is [(5/6)^2 * (1/6)] + [(5/6)^4 * (1/6)] + [(5/6)^6 * (1/6)] + ... = (25/36) * (1/6) / (1 - (25/36)) = 25/66.

Therefore, the probability that 2 appears in an even number of throws is P(E) = P(E1) + P(E2) = 5/36 + 25/66 = 15/66 + 25/66 = 40/66 = 20/33.