"We can just do "(3/5)*80 = 48". If you want a proof then"
"read further here underneath."
P["Kristen hits k times on 80"] = C(80,k) (3/5)^k (2/5)^(80-k)
"with "C(n,k) = (n!)/((n-k)!*(k!)) " (combinations)"
"(binomial distribution)"
"Expected value = average = E[k] :"
sum_{k=0}^{k=80} k*C(80,k) (3/5)^k (2/5)^(80-k)
= sum_{k=1}^{k=80} 80*(79!)/((80-k)! (k-1)!) (3/5)^k (2/5)^(80-k)
= 80*(3/5) sum_{k=1}^{k=80}C(79,k-1) (3/5)^(k-1) (2/5)^(80-k)
= 80*(3/5) sum_{t=0}^{t=79} C(79,t) (3/5)^t (2/5)^(79-t)
"(with "t=k-1")"
= 80*(3/5)*1
= 48
"So for a binomial experiment, with "n" tries, and probability"
p" for the chance of success on a single try, we have in general"
"expected value=average= "n*p" (of the number of successes)"
