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 \( e^{-2}=0.1353 \)]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.1804
For the Poisson approximation, the mean is \( \lambda=np \), with p = 0.1% = 0.001:
\( \displaystyle \lambda=2000\times 0.001=2 \)
\( \displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \)
\( \displaystyle P(X=3)=\frac{e^{-2}\,2^3}{3!} \)
\( \displaystyle =\frac{0.1353\times 8}{6}=0.1804 \)