In a factory manufacturing certain spare parts, 2% of the output is found defective. If spares are packed in 200 units, the probability that there are 4 defective spares in a pack is
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a
64e−4
b
32e−4
c
643e−4
d
323e−4
answer is D.
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Detailed Solution
p=2100=150 λ=np=200.150=4 P(X=4)=e−4444!=323e−4 the probability that there are 4 defective spares in a pack is = 323e−4