There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Let there be x number of defective items in a sample of ten items drawn successively.

Now, as we can see that the drawing of the items is done with replacement. Thus, the trials are Bernoulli trials.

Now, probability of getting a defective item,

we can say that x has a binomial distribution, where

Thus, P(X = x) = nCxqn-xpx , where x = 0, 1, 2, …n

Probability of getting not more than one defective item = P(X ≤1)

= P(X = 0) + P(X = 1)

= 10C0 (19/20)10(1/20)0 +10C1 (19/20)9(1/20)1