Question

In a certain factory making razor blades there is a small probability of 1/500 for any blade
to be defective in each packet. The blades are supplied in packets of 10. Calculate the no.
of packets having a) no defective blade b) 1 defective blade c) 2 defective blade in a
consignment of 10,000 packets.

Answer 1

p=0.002

L=10*0.002=0.02

P(n)=(Lⁿ/n!)*e^-L

a) P(0)=(0.02⁰/0!)*e^-0.02=0.98

N₀=0.98*10000=9800

b)P(1)=(0.02¹/1!)*e^-0.02=0.0196

N₁=0.0196*10000=196

c)P(2)=(0.02²/2!)*e^-0.02=0.000196

N₂=0.000196*10000 =2