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We roll a fair die repeatedly. Let X denote the number of rolls needed until seeing the first 6, and Y the number of rolls needed to see the second 6 (counted from the first roll). Find the probability of P=(Y=2X+a) where a=3.

User Bskool
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Final answer:

To find the probability of P=(Y=2X+a) where a=3, we need to understand the probability distribution of X and Y. X represents the number of rolls needed to see the first 6, and Y represents the number of rolls needed to see the second 6. Using the geometric distribution, we can calculate the probability of P=(Y=2X+a) = P(Y=5) = (5/6)^4 * (1/6).

Step-by-step explanation:

To find the probability of P=(Y=2X+a) where a=3, we need to understand the probability distribution of X and Y.

Since X represents the number of rolls needed to see the first 6, it follows a geometric distribution with parameter p=1/6 (since there is a 1/6 probability of rolling a 6 on each roll).

Similarly, Y represents the number of rolls needed to see the second 6, counting from the first roll. Since the first 6 already occurred on the first roll (X=1), we can think of Y as the number of additional rolls needed to see the second 6, which also follows a geometric distribution with parameter p=1/6.

Now, let's calculate the probability of P=(Y=2X+a), where a=3:

First, let's find the probability P(X=1). Since X follows a geometric distribution, P(X=1) = p = 1/6.

Next, let's find the probability P(Y=2X+a) = P(Y=2(1)+3) = P(Y=5).

Since Y follows a geometric distribution with parameter p=1/6, we can calculate P(Y=5) = (1-p)^4 * p = (5/6)^4 * (1/6).

Therefore, the probability of P=(Y=2X+a) = P(Y=5) = (5/6)^4 * (1/6).

User IvanRF
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