# Homework Help: Independence and conditional probability

1. Jul 20, 2006

### island-boy

if X and Y are events which are independent of each other, but neither are independent with A,

is this equality true for conditional probabilities:
P( X, Y | A) = P(X|A) * P(Y|A)

if not,
how do you solve for P(A | X,Y)
given that you only know P (A) and P(X|A) and P(Y|A)?

The reason I came up with the above probability where I have:
$$P(A| X, Y) = \frac {P(X, Y | A) P(A)}{P(X, Y |A) P(A) + P(X, Y | A^c) P (A^c)}$$
is that I used Baye's Thm.

Note: P(X, Y |A) is not given.

2. Jul 20, 2006

### HallsofIvy

First, by P(X,Y) do you mean P(X and Y) or do you mean P(X or Y)?

Assuming that you mean P(X and Y) then, yes, since X and Y are independent, P(X)= P(Y) so P(X and Y| A)= P(X|A)P(Y|A).

3. Jul 20, 2006

### island-boy

Cheers for the help, HallsofIvy.

well, this is how it is stated in the problem:
X is the event of testing negative in a drug test
Y is the event of testing positive in the drug test
A is the event of having a disease.

Given that a person went to have a drug test three times, testing positive once and negative twice. What is the probability he has a disease?

here, X and Y are mutually exclusive events. However, are $$X_1$$ and $$X_2$$ mutually exclusive? It would seem that testing thrice would mean the question is asking for the union of the 3 events. Is this correct?

so would the solution be
$$P(A| X, X, Y) = \frac {P(X|A)P(X|A)P(Y|A)P(A)}{P(X|A^c)P(X|A^c)P(Y|A^c)P(A^c)}$$ ?

Last edited: Jul 20, 2006