The data follow a Binomial.
a. Expected number of good items
E(X)=nP
n= 10 and P= 0.8
E(X)=10×0.8=8
b . Probability of no defective.
P(=0) can be obtained from =BINOM.DIST(0,10,0.2,FALSE) Excel formula which yields 0.107374.
Similarly, substituting to the Binomial distribution, we have (1−P)n=0.810=0.107374.
c. P(≥3 defectives).
The Excel formula =1-BINOM.DIST(2,10,0.2,TRUE) can be used which yields 0.3222.
Similarly from Binomial distribution, 1−(P(=0)+P(=1)+P(=2)) can be obtained:
P(=0)=0.107374 from (b),
P(=1)=(110)×0.2×0.89=0.268435 ,
P(=2)=(210)×0.22×0.88=0.30199 .
Thus, P(≥3)=1−0.107374−0.268435−0.30199=0.3222 .
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