# Conditional probability with joint condition

1. Oct 11, 2014

### Cognac

So say I have Pr(Z|X&Y)

I'm guessing that it follows the standard Pr(A|B)=[Pr(B|A)Pr(A)]/Pr(B)

So Pr(Z|X&Y)=[Pr(X&Y|Z)Pr(Z)]/Pr(X&Y)?

Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?

Last edited: Oct 11, 2014
2. Oct 11, 2014

### FactChecker

What? If A=B, this would say that Pr(A)=1.
No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.

Last edited: Oct 11, 2014
3. Oct 11, 2014

### Cognac

Thanks, that example about independence makes things make more sense now.

Oh sorry about that. I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?

4. Oct 12, 2014

### FactChecker

By "divide by the conditional", I assume you mean "divide the right side by P(A)". Yes, Bayes' Rule always works. The only exception to the form of Bayes' Rule that you are using is if P(A) = 0. Then both P(B|A) and the division of the right side by 0 are problems. (Some statements of Bayes' Rule talk about a partitioning of the space. If you use ever that form, make sure that condition is met.)