Question

Cycle tyres are supplied in lots of $10$ and there is a chance of $1$ in $500$ to be defective. Using poisson distribution, the approximate number of lots containing no defectives in a consignment of $10,000$ lots if $e^{-0.02}=0.9802$ is

• A. $9980$
• B. $9998$
• C. $9802$
• D. $9982$

Answer 1

Answer: (C)

$9802$

Here $\lambda =\frac{1}{500}\times 10=0.02$
Thus probability that lot is not defective is $=P(X=0)=\frac{e^{-0.002}({.0020}^0}{0!}=e^{-.002}=0.9802$
Hence number of no defective lots out of $10,000$ is $=.9802\times 10,000=9802$