#"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)"#