Suppose you cast and keep casting a die until you get either number 5 or 6? What is the probability that you will stop after even number of casts?
Hint: Use series to solve.
To find the probability of stopping after an even number of casts, let's break the problem down by considering the probability of stopping after each round.
We can represent the outcomes of each round as a geometric series. In the first round, the probability of not rolling a 5 or 6 is 4/6, as there are 4 outcomes (1, 2, 3, 4) out of 6 total outcomes. Hence, the probability of stopping after the first round is 2/6, which means we roll a 5 or 6.
In the second round, the probability of not rolling a 5 or 6 is (4/6) * (4/6), as for each round, the probability of rolling a number other than 5 or 6 remains constant. Therefore, the probability of stopping after the second round is 2/6 * 2/6 = 4/36.
We can continue this pattern for subsequent rounds, where the probability of stopping after each round is given by (2/6)^(n-1), where n is the round number.
Since we are interested in stopping after an even number of casts, we need to sum the probabilities of stopping after rounds 2, 4, 6, etc., up to infinity.
To find the sum of an infinite geometric series with a common ratio less than 1, we can use the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio.
In our case, a = 4/36 and r = (2/6)^2 = 1/9.
So, substituting these values into the formula, we get:
S = (4/36) / (1 - 1/9)
= (4/36) / (8/9)
= (4/36) * (9/8)
= 1/9 * 1/2
= 1/18
Therefore, the probability of stopping after an even number of casts is 1/18.