# Thread: A question on: Taking 2 fail products of 20

1. ## 20 items, 8 are failed, probability to coose 2 failed items..

Hello,

Sorry for not being clear I am editing my question, I myself had not read it thoroughly, (was tired at the time)...

This is the exercise:

There are 20 items in total. Among them there are 8 failed items. We choose randomly 2 items. What is the proability that these 2 items are both failed.

Soltion (in the book):

Threre are C(20,2) ways we can choose 2 any items.There are C(18,2) ways we can take 2 failed items. And the probability is C(18,2)/C(20,2).

I am sure this is trivial exercise containing a typographical error, and we should have C(8,2) insted of C(18,2), right?

Thank you very much!

My aplogies again for not being clear!

2. ## Re: A question on: Taking 2 fail products of 20

Originally Posted by Melsi
Hello,

We choose 2 products of 20, this is C(20,2) different ways (this is clear).

Then for both products being failed there are C(18,2) ways we can choose. This one I don't get it!?

Solution= C(18,2)/C(20,2) Why??

In addition I can understand very well this problem, choosing two asses from 52 cards = C(4,2)/C(52,2).

Can the last problem help me somehow understand the first one, do they relate in any way?

Thank you very much!
Post the whole question. We are not mind readers, although if I had to attempt it I'd guess that the question is something like: 18 of 20 products are faulty. You choose 2. Find the probability that both are faulty.

But the fact is that we shouldn't have to be mind readers. If you want help, you're meant to make it easy for people to help ....