Question

A company knows on the basis of past experience that $2$% of its blades are defective. The probability of having $3$ defective blades in a sample of $100$ blades if $e^{-2}=0.1353$ is

• A. $0.1353$
• B. $0.1804$
• C. $0.2706$
• D. $0.3606$

Answer 1

Answer: (B)

$0.1804$

Here P.D parameter $\lambda =\frac{2}{100}\times 100=2$
Hence number of probability that 3 blade are defective is $=P(X=3)=\frac{e^{-2}(2{)}^3}{3!}=\frac{4}{3}{e}^{-2}=\frac{4}{3}\times .1353=.1804$