# A EASY QUESTION FOR probability probability A easy question for probability

1. Aug 7, 2012

### fhjop1

Let X; Y be two random variables with the following joint distribution（in the attachment）:

(a) Compute p(X) and p(Y )
(b) Compute p(X|Y = 1)
(c) compute p(Y = 1|X = 1) using (a) and (b)
(d) Are X and Y statistically independent? Show all your working.

This is my answer, but I don't know whether it's correct
(a) joint probabilities:
p(X=1)= 0+1/8=1/8
p(X=2)= 3/4+1/8=7/8
so p(X)= 1/8+7/8=1
p(Y=1)= 0+3/4=3/4
p(Y=2)=1/8+1/8=1/4
so p(Y)=3/4+1/4=1
just like Millennial said, I also don't understand this question, because I think it doesn't make any sense.

(b) p(Y=1,X) = p(Y=1)p(X|Y=1)
so p(X|Y=1) = p(Y=1, X)/p(Y=1) =?/(3/4)
here I have a question, I don't know how to compute p(Y=1,X), and btw does p(Y=1,X) = p(X, Y=1)?

(c) because I can't actually compute (b), so just leave it behind for now.

(d) because p(X=1,Y=1) = 0 and it doesn't equal to p(X=1)p(Y=1) = 1/8,so they are not statistically independent.

#### Attached Files:

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Last edited: Aug 7, 2012
2. Aug 7, 2012

### Millennial

I didn't understand anything. Are X and Y, 1 or 2? Or is 1 the happening probability of Y and X?

3. Aug 7, 2012

### clamtrox

You misunderstood the assignment; it's you who are supposed to do it, not us. Maybe show us what you got so far and we might be able to help.

4. Aug 7, 2012

### Staff: Mentor

The table may be clearer if the label p(X,Y) didn't appear on the figure, but instead were printed underneath the figure as a caption.

5. Aug 7, 2012

### Simon Bridge

fhjop1: welcome to PF.
clamtrox is correct - since you say the question is "easy", what's the problem?