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%