# Expected values for random variables

1. Sep 19, 2011

### cue928

I am stuck on the following problem: Five items are to be sampled from a large lot of samples. The inspector doesn't know that three of the five sampled items are defective. They will be tested in randomly selected order until a defective item is found, at which point the entire lot is rejected. Y is the number of firing pins the inspector must test. Find, graph the probability distribution of Y.

I understand that p(1) = 3/5. But it is saying that p(2) =3/10 and p(3)=1/10, I do not see how they are calculating that.

2. Sep 19, 2011

### CompuChip

So p(2) is the probability that the second sample tested is defective. That means that you are looking at the case where the first one is non-defective (otherwise testing would stop there!) and the second one is broken.
What are the probabilities for each of those? Then how do you calculate p(2)?

3. Sep 19, 2011

### cue928

That's what I can't figure out. Is p(2) asking what the chance of the second draw being defective?

4. Sep 19, 2011

### CompuChip

It says "Y is the number of firing pins the inspector must test." and that they will stop testing the moment a defective item is found.

p(2) is "sloppy" shorthand for P(Y = 2), that is: when you do the experiment, what is the probabillity of finding the value 2 for Y.

5. Sep 19, 2011

### Ray Vickson

After finding that the first item is non-defective, how many items are left to test? How many of those are defective?

RGV