"P(x)=\\frac {n!} {(n-x)!x!}p^xq^{n-x}"
where "n=5, p=1\/5,q=1-p=4\/5"
So,
"P(3)=\\frac {5!} {(5-3)!3!}(\\frac 1 5)^3(\\frac 4 5 )^{5-3}=0.08*0.64=0.0512=5.12\\%"
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