The manager at MMM Enterprise has come up with a double samp

The manager at MMM Enterprise has come up with a double sampling procedure to ensure quality.

First, he selects a lot size of 40 and takes 4 samples from it. He accepts the lot if at most one article is defective. The lot is rejected if at least 3 articles turn out to be defective. If 2 are defective, a second sample of one article is taken.

The lot is rejected if the article in the second sample is defective. He accepts the lot when the article in the second sample is good.

            (a) Compute the probability of acceptance if 25 percent of lot is defective.

Solution

Number of defective items= 0.25*40= 10

Number of good items= 30

The lot is accepted either in the first go or in the second go :

Probability of getting accepted in the first go =probability that at most one item is defective
= probability that one or none item is defective

= 4C1/10C1 * 4/40 = 0.04

Probability of getting accepted in the second go = prob. That 2 are defective in the first go and the one in the second go is good

= 10C2/ 40C2 * 30/40 = 45/780*3/4= 0.0432

Thus, required probability = 0.04+0.0432= 0.0832