Let X= the number of defective items: X∼Bin(n,p)
P(X=x)=(xn)px(1−p)n−x Given that p=0.20,n=10
a. The expected number of good items
E(XC)=n(1−p)=10(1−0.2)=8
b. Probability that none is defective
P(X=0)=(010)0.20(1−0.2)10−0=0.810=0.1073741824 c. Probability that at least 3 are defective
P(X≥3)=1−P(X=0)−P(X=1)−P(X=2)=
=1−(010)0.20(1−0.2)10−2−(110)0.21(1−0.2)10−1−
−(210)0.22(1−0.2)10−2=1−0.1073741824−
−0.268435456−0.301989888=0.3222004736≈
≈0.3222
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