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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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$64 e^{- 4}$

$32 e^{- 4}$

$\frac{64}{3} e^{- 4}$

$\frac{32}{3} e^{- 4}$

answer is D.

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Detailed Solution

$p = \frac{2}{100} = \frac{1}{50}$

$λ = n p = 200. \frac{1}{50} = 4$

$I n p o i s s i o n d i s t r i b u t i o n P (X = r) = \frac{e^{- λ} λ^{r}}{r!} P (X = 4) = e^{- 4} (\frac{4^{4}}{4!}) = \frac{32}{3} e^{- 4}$

the probability that there are 4 defective spares in a pack is = $\frac{32}{3} e^{- 4}$

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