# Conditional Probability for discrete random variables.

1. Apr 24, 2012

### XodoX

1. The problem statement, all variables and given/known data
Compute P(X=k l X+Y=p)

2. Relevant equations

3. The attempt at a solution

No idea. Kind of understand page #1. Although it seems like there's a lot of unnecessary stuff. Could have gone straight from the top to the bottom. And I don't know why/if you even have to substitute the X+Y=p for Y=k-p. Totally lost on page 2. No idea whats going on there. Says it's being split up because it's independent, but no idea where the 1/3 and 2/3 for 0 and 1 come form. Let alone the rest of page 2. :uhh:
So, in short: What's exactly step 1,2,3 etc. ?

#### Attached Files:

• ###### Problems.pdf
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2. Apr 24, 2012

### Ray Vickson

The 1/3 and 2/3 for 0 and 1, etc., are just examples, so the writer has some definite numbers to work with when practicing use of the formulas.

RGV

3. Apr 25, 2012

### mathmajor23

I'm also trying to solve this similar problem, and also have no idea how to go about solving it.

4. Apr 25, 2012

### Ray Vickson

For two events A and B we have
$$P(A|B) = \frac{P(A \cap B)}{P(B)}.$$

That's all there is to it. Just figure out what are the events A and B in your problem.

RGV

Last edited: Apr 25, 2012