Let X= the number of defective fuses: X∼B(n,p)
P(X=x)=(xn)px(1−p)n−x Given n=6,p=0.05
(a) What is the probability one fuse will be defective?
P(X=1)=(16)(0.05)x(1−0.05)6−1≈0.232134 (b) What is the probability at least one fuse will be defective?
P(X≥1)=1−P(X=0)=
=1−(06)(0.05)0(1−0.05)6−0≈0.264908(c) What is the probability that more than one fuse will be defective, given that at least one is defective?
P(X>1,X≥1)=P(X>1)=P(X≥1)−P(X=1)≈
≈0.264908−0.232134≈0.032774
P(X>1∣X≥1)=P(X≥1)P(X>1,X≥1)≈
≈0.2649080.032774≈0.1237
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