Suppose that three identical parts are chosen for inspection

Suppose that three identical parts are chosen for inspection. Each part can be defective with probability p independently of the other parts. Parameter p is, in turn, a uniform random variable over the interval (0, 1]. What is the probability that exactly two parts are defective?

Solution

Let defective be represented as D and effective as E

So, 3 parts can be EEE,EED,EDE,EDD,DEE,DED,DDE,DDD

Out of these, exactly 2 are defective in EDD, DED, DDE

So, 3 out of 8 cases have exactly 2 parts defective.

Probability for EDD = (1-p)*p*p

Probability for DED = p*(1-p)*p

Probability for DDE = p*p*(1-p)

So, final answer is 3*p*p*(1-p) = 3p^2*(1-p)