This is a binomial distribution problem. Probability of success on single trial

(a head on one toss) p=0.5. Number of trial n=5, number of successes of

interest k=4.

Then the required probability is:

f(k;n,p)=C(n,k) p^k (1-p)^(n-k)

which in this case is:

f(4;5,4)=5 (0.5)^4 (0.5)=0.15625

or about 15.6%.

RonL