A fair die is thrown 20 times. The probability that on the $10^{th}$ throw, the fourth six appears is

This is an application of the negative binomial distribution, where we need exactly 3 sixes in the first 9 throws and a six on the 10th throw. The probability of exactly 3 sixes in 9 throws is 9C3 * (1/6)^3 * (5/6)^6. Multiplying this by the 1/6 probability of getting a six on the 10th throw gives 9C3 * (1/6)^4 * (5/6)^6 = 84 * 5^6 / 6^10.

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