Mean number of heads = 0*P(0 heads) + 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads). We don't need the probability of 0 heads since that will be multiplied by 0. So: Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads). The probability of getting x heads out of n tosses of a fair coin isUse that formula to fill in this table of probabilities: P(1 head) P(2 heads) P(3 heads) ball#1(1 toss) 1/2 0 0 ball#2(2 tosses) 1/2 1/4 0 ball#3(3 tosses) 3/8 3/8 1/8 Now we need to get P(1 head), P(2 heads), P(3(heads) P(1 head) = P[(ball#1 AND 1 head) OR (ball#2 AND 1 head) OR (ball#3 AND 1 head)] Using the rule of "AND implies multiplication" and "OR implies addition": P(1 head) = P(2 heads) = P[(ball#2 AND 2 heads) OR (ball#3 AND 2 heads)] = P(3 heads) = P[ball#3 AND 3 heads] = So, Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads) = . So, surprisingly, the mean number of heads is 1 head!!! Edwin