Trouble with a conditional probability problem

Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

So I'm having some trouble with the version of this problem that requires replacement.

Let, E be the event that the first and third draws are both white balls, and F be the event that the sample contains exactly 3 white balls.

So I'm looking for the conditional probability of E given F which is;

Where the intersection is the probability of drawing 1 non-white ball and 3 white balls. And since there are two valid permutations (wwwn,wnww) we introduce a factor of two

My problem is with the probability of F; which I'm defining as the odds of drawing one non-white ball and 3 white balls. I'm getting an answer that is less then the intersection of E and F so it's clearly nonsense.

Do I need to permute the balls in F or something?

Re: Trouble with a conditional probability problem

Quote:

Originally Posted by

**bkbowser** Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

Let's look at the condition: "given that the sample drawn contains exactly 3 white balls".

That means we can cut down on the space to these events.

BWWW, WBWW, WWBW, WWWB

How many of those have a white in the first and third place?

Re: Trouble with a conditional probability problem

Quote:

Originally Posted by

**Plato** Let's look at the condition: "given that the sample drawn contains exactly 3 white balls".

That means we can cut down on the space to these events.

BWWW, WBWW, WWBW, WWWB

How many of those have a white in the first and third place?

Just the two. I'm not sure where this is going unless I just need to double that again?

http://www.texify.com/img/%5CLARGE%5...%7D%29%5E3.gif

?

Re: Trouble with a conditional probability problem

Quote:

Originally Posted by

**Plato** Let's look at the condition: "given that the sample drawn contains exactly 3 white balls".

That means we can cut down on the space to these events.

BWWW, WBWW, WWBW, WWWB

How many of those have a white in the first and third place?

Oh no the conditional probability must be one half.

So then how do I show that using the conditional probability formula?