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If 0.1% of the bolts manufactured by a machine are found to be defective. Find the probability that, in a sample of 2000 bolts chosen at random, exactly 3 will be defective using the Poisson distribution. [Given e−2=0.1353e^{-2} =0.1353]

Solution

✅ Correct Option: 3

Here λ=np=2000×0.001=2\lambda=np=2000\times0.001=2. Then P(3)=e−2233!=0.1353×86≈0.1804P(3)=\frac{e^{-2}2^3}{3!}=\frac{0.1353\times8}{6}\approx0.1804.

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