If 8 defective and 12 nondefective items are inspected one by one, at random and without replacement, what is the probability that (a) the first four items inspected are defective; (b) from the first three items at least two are defective?

If 8 defective and 12 nondefective items are inspected one by one, at random and without replacement, what is the probability that (a) the first four items inspected are defective; (b) from the first three items at least two are defective?

Be aware that you should not get more than one account per person. This said, back to the question.

You should draw a probability tree to help you.

I'll call 'D' defective, and 'N' non defective. The order to which they are written is 1st to last from left to right in my notations.

(a) P(DDDD) = P(1st D) x P(2nd D) x P(3rd D) x P(4th D)

(b) P(2 defective or 3 defective in first three) = P(2 defective) + P(3 defective) = [P(DDN) + P(DND) + P(NDD)] + P(DDD)

Do you understand?

In any case, reply.