A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws , is
Question
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is
Solution
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:
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E1: 2 appears in the second throw. The probability of this event is (5/6) * (1/6) = 5/36.
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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.
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