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

In the first 9 throws, we should have three sixes and six non-sixes; and a six in the 10th throw, and thereafter it does not matter whatever face appears. So, the required probability is 9C3163×566×16×1×1×1×⋯×110 times=84×56610