1.Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B are the next events: A: "Four items are non-defective" B: "two items are defective and two items are non-defective Find the probability of events A and B.

(Sorry for bad english)

Okay, i figured out that we can solve 1.(A) by using combinatorics formula, so it will be 16C4/11C4 which equals 330/1820=33/182=0.181 i guess this is the right answer, but how do i solve b?

First, here is a better version in English of this problem as I understand it:

Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B *be* the *following* events: A= "Four items are non-defective", B= "two items are defective and two items are non-defective. Find the probability of *each of the* events A and B.

You are saying that Pr(A) = 16C4/11C4; I think you meant 11C4/16C4 = 330/1820. This is the number of ways to chose 4 of the 11 non-defective items, over the number of ways to choose any 4 items. That is correct.

To find Pr(B) by the same technique, your numerator has to be the number of ways to choose 2 from the 5 defective

**and** 2 from the 11 non-defective. Can you see how to do that? Make an attempt.