Answer to Question #324561 in Statistics and Probability for Vennela

Question

The probability that a blade manufactured by a factory is defective is 1/500. Blades are

packed in packets of 10 blades. Find the expected number of packets containing (i) no

defective blade (ii) one defective blade (iii) 2 defective blades, in a consignment of 10000

packets.

How can we solve this by using only poison distribution

Answer 1

p=$\frac{1}{500}$ =0.002

per packet =0.002$\cdot$ 10=0.02

Thus $\lambda$ =0.02

P(X)=$\frac{\lambda^xe^{-\lambda}}{x!}$

P(0)=$\frac{0.02^0}{0!}\cdot e^{-0.02}=0.98$

The approximate # of packets with no defective blades is 0.98*10000=9800

P(1)=$\frac{0.02^1}{1!}\cdot e^{-0.02}=0.0196$

The approximate # of packets with 1 defective blades is 0.0196*10000=196

P(2)=$\frac{0.02^2}{2!}\cdot e^{-0.02}=0.0000196$

The approximate # of packets with 2 defective blades is 0.0000196*10000=2