Let "X=" the number of defective items: "X\\sim Bin(n, p)"
"P(X=x)=\\binom{n}{x}p^x(1-p)^{n-x}" Givven that "p=0.2, n=10"
a. The expected number of good items
"P(\\bar{X})=n(1-p)=10(1-0.2)=8" b. Probability that none is defective
"P(X=0)=\\binom{10}{0}(0.2)^0(1-0.2)^{10-0}=""(0.8)^{10}=0.1073741824\\approx0.1074" c. Probability that at least 3 are defective
"P(X\\geq3)=1-P(X<3)=""=1-P(X=0)-P(X=1)-P(X=2)=""=1-\\binom{10}{0}(0.2)^0(1-0.2)^{10-0}-""-\\binom{10}{1}(0.2)^1(1-0.2)^{10-1}-\\binom{10}{2}(0.2)^2(1-0.2)^{10-2}=""1-(0.8)^{10}-10(0.2)(0.8)^9-45(0.2)^2(0.8)^8=""=1-0.1073741824-0.268435456-""-0.301989888=0.3222004736\\approx""\\approx0.3222"
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