Question

) An electronics retailer purchases ICs from a manufacturer. The manufacturer indicates that the defective rate of the ICs is 3%. a. An inspector randomly picks 20 ICs from a shipment. What is the probability that there will be at least one defective IC in the shipment? b. Suppose the retailer receives 10 shipments in a month and the inspector randomly tests 20 ICs per shipment. What is the probability that there will be exactly 3 shipments each containing at least one defective IC in the 20 that are selected and tested from each shipment?

Answer 1

Solutions:-

Given information here is Bionomial distrubution.

for bionomial distribution with

p = probability of success.

n = number of trials.

we have

P(X=x) = xCn*p^x*(1-p)^(n-x)

using this formula

a)

here

p= 0.03, n=20

P(at least one defective item among these 20) = P(X>=1) = 1-P(X=0)

= 1-0C20*0.03^0*(1-0.03)^(20-0)

=0.456

b)

here , p=P(at least one defective item among these 20)= 0.456, n=10

required prob. = P(X=3) =3C10*0.456^3*(1-0.456)^(10-3)

= 0.16

Conclusion:-

Answer part (A) = 0.456

Answer part (B) = 0.16