### Instruction: We roll a fair 6-sided die 5 times.  What is the probability that we get a 6 in at most 2 of the rolls?### Response: This is a binomial probability problem, where we have a fixed number of independent trials (5 rolls), each with a constant probability of success (getting a 6, which is 1/6) or failure (not getting a 6, which is 5/6).
The probability of getting exactly k successes in n trials is given by the binomial formula: (n choose k) * p^k * (1-p)^(n-k), where p is the probability of success and (n choose k) is the number of ways to choose k objects out of n, which is n!/((n-k)!*k!).
In this problem, we want the probability of getting a 6 in at most 2 of the rolls, which means we can have 0, 1, or 2 successes.
So we need to add the probabilities of these three cases.
The probability of getting 0 successes (no 6's) is (5 choose 0) * (1/6)^0 * (5/6)^5, which simplifies to 1 * 1 * (5/6)^5, which is about 0.4019.
The probability of getting 1 success (one 6) is (5 choose 1) * (1/6)^1 * (5/6)^4, which simplifies to 5 * (1/6) * (5/6)^4, which is about 0.4019 as well.
The probability of getting 2 successes (two 6's) is (5 choose 2) * (1/6)^2 * (5/6)^3, which simplifies to 10 * (1/36) * (125/216), which is about 0.1608.
Adding these three probabilities, we get 0.4019 + 0.4019 + 0.1608, which is about 0.9646.
This is the probability of getting a 6 in at most 2 of the rolls.