Question

A company is shipped computers. The company randomly tests 10 computers from each shipment (with replacement). If at most 1 computer is defective in the sample, the company accepts the shipment. The manufacturer sends 4184 computers with a known 2.61% rate of defective computers. What is the probability accurate to 6 decimal places that the shipment is rejected?

Answer 1

let

p = probability of defective computers = 2.61% = 0.0261. = prob. of success.

n = number of computers randomly selected (with replacement) = 10.

X : number of computers are defective in 10 computers.

X takes values 0 ,1, 2, .........., 10.

Since prob. of success ( prob. of a computer is defective) is constant for each selection and selection is independent to each other.

Hence X follows binomial distribution with parameter n = 10 and p = 0.0261.

Comapny accepts the shipment if number of defectives is atmost 1.

i.e. if X < =1.

Hence Company rejected the shipment if X > 1.

P ( Shipment is rejected ) = P ( X > 1) = 1 - { P(X=0) + P(X=1)}