Question

I have a batch of 100 memory chips, each is defective with probability 0.05, independently of the others. Use (i) the binomial distribution and (ii) the Poisson distribution to approximate the probability that there are fewer than three defective chips in the batch. Give your answers correct to three decimal places

Answer 1

Ans:

Given that

n=100,p=0.05

i)Using binomial distribution:

P(x<3)=P(x=0)+P(x=1)+P(x=2)

=100C₀*0.05⁰*0.95¹⁰⁰+100C₂*0.05¹*0.95⁹⁹+100C₂*0.05²*0.95⁹⁸

=0.0059+0.0312+0.0812

=0.1183 $\approx 0.118$

ii)Poisson approximation:

mean=np=100*0.05=5

P(x<3)=P(x=0)+P(x=1)+P(x=2)

=e^-5*(5^0/0!+5^1/1!+5^2/2!)

=e-5*(1+5+12.5)=e^-5*18.5=0.125