Question

A manufacturer who produces medicine bottles finds that $0.1$% of the bottles are defective. The bottles are packed in boxes containing $500$ bottles. A drug manufacturer buys $100$ boxes from the producer of bottles. Using poisson distribution,the number of boxes with at least one defective bottle is

• A. $100(1-e^{-0.1})$
• B. $100(1-e^{-0.5})$
• C. $100(1-e^{-0.05})$
• D. $100(1-e^{-0.01})$

Answer 1

Answer: (B)

$100(1-e^{-0.5})$

Here $\lambda =\frac{0.1}{100}\times 500=0.5$
Hence number of boxes out of 100 which contain at least one defective bottle is,
$=100(1-P(X=0\left)\right)=100(1-e^{-0.5})$