A die is rolled until a 3 occurs. By conditioning on the outcome of the irst roll,ind the probability that an even number of rolls is needed.

To solve this problem, we need to consider two scenarios: one where the first roll is a 3, and one where it is not.

1. If the first roll is a 3, then an odd number of rolls is needed (just 1 roll), so this does not contribute to the probability we are trying to find.

2. If the first roll is not a 3, then we need to continue rolling. There are 5 possible outcomes that are not a 3 (1, 2, 4, 5, 6), each with a probability of 1/6. For each of these outcomes, we are essentially starting the problem over, but with one roll already completed. This means we are now looking for the probability that an odd number of additional rolls is needed, which is the same as the original problem (the probability that an even number of rolls is needed).

Therefore, we can set up the following equation to solve for P(E), the probability that an even number of rolls is needed:

P(E) = 5/6 * (1 - P(E))

Solving for P(E), we get:

P(E) = 5/11

So, the probability that an even number of rolls is needed to roll a 3 is 5/11.