Question

Part B: Answer the following questions based on special probability distributions.

Q11. It is known that fifteen percent of Printed Circuit Board (PCB) fabricated at the PCB lab are defective. If 5 PCBs are fabricated, find the probability that

a) None of the PCB are defective

b) Fewer than two PCB are defective

c) More than three PCB are defective

Answer 1

Given that it is known that fifteen percent of Printed Circuit Board (PCB) fabricated at the PCB lab are defective.

Hence probability that a Printed Circuit Board (PCB) fabricated at the PCB lab is defective is p=15%=0.15

Now 5 PCB's are fabricated.

Let,

X=Number of defective PCB's out of the 5 fabricated PCB's

Binomial Distribution

A random variable X is said to have a binomial distribution if its PMF(Probability Mass Function) is given by,

0<p<1.

Notation: X~Binomial(n,p)

Coming back to our problem

Let,

X=Number of defective PCB's out of the 5 fabricated PCB's

(a) Here we need to find the probability that none of the PCB are defective.

Hence the probability that none of the PCB are defective is 0.4437

(b) Here we need to find the probability that fewer than two PCB are defective.

Hence the probability that fewer than two PCB are defective is 0.8352

(c) Here we need to find the probability that more than three PCB are defective.

Hence the probability that more than three PCB are defective is 0.0022