Let X= the number of defective items: X∼Bin(n,p)
P(X=x)=(xn)px(1−p)n−x Givven that p=0.2,n=10
a. The expected number of good items
P(Xˉ)=n(1−p)=10(1−0.2)=8 b. Probability that none is defective
P(X=0)=(010)(0.2)0(1−0.2)10−0=(0.8)10=0.1073741824≈0.1074 c. Probability that at least 3 are defective
P(X≥3)=1−P(X<3)==1−P(X=0)−P(X=1)−P(X=2)==1−(010)(0.2)0(1−0.2)10−0−−(110)(0.2)1(1−0.2)10−1−(210)(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≈≈0.3222
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