Your question is related to the binomial distribution. The question doesn't mention if this is a fair coin. So I assume the probability of getting ahead in a single toss = P. As a result, the probability of getting a tail in a single toss = (1-P).The probability of getting 22 heads and 28 tails = C(22,28)*(P^22)*[(1-P)^28]Where C(x,y) is the combination calculator/operator. The idea can be substantially generalized. If we toss a coin n times, and the probability of a head on any toss is P (which need not be equal to 1/2, the coin could be unfair), then the probability of exactly k heads