If two fair dice are rolled 10 times, what is the probability of at least one 6 (on either die) in exactly five of these 10 rolls?

About 10.84% is the answer. The exact value is (5^10) 7 (11^5) / (6^18) First, determine the probability that you'll get at least one 6 on a single throw of two dice. Of the 36 ways that two dice can fall, eleven of them have at least one 6. So the probability is 11/36 for a single throw. Now what's the probability of exactly 1 throw out of 10 having at least one 6? That would be + 11/36 * (25/36) ^9 + 25/36 * 11/36 * (25/36) ^8 + (25/36) ^2 * 11/36 * (25/36) ^7 ... + (25/36) ^9 * 11/36 or 11/36 * (25/36) ^9 * 10 In general the probability of exactly n events out of m having a condition with probability p happen will be p^n * (1-p) ^ (m-n) multiplied by the number of ways you can select n out of m items. And since the formula for selecting n of m items is m! / (n! (m-n) !), the overall formula becomes m! / (n! (m-n) !) p^n * (1-p) ^ (m-n) so 10! / (5!5!) * (11/36) ^5 * (25/36) ^5 = 252 * (161051/60466176) * (9765625/60466176) = (5^10) 7 (11^5) / (6^18) ~ 0.108402426 So the exact answer is (5^10) 7 (11^5) / (6^18) and the approximate answer is 0.108402426 which is about 10.84%