# Calculating Joint Distribution

1. Oct 18, 2009

### BlueScreenOD

1. The problem statement, all variables and given/known data
Let X and Y be two random variables, with joint distribution given by the following table:
Code (Text):
a
b       1       2       3       4
--------------------------------------
1       16/136  3/136   2/136   13/136
2       5/136   10/136  11/136  8/136
3       9/136   6/136   7/136   12/136
4       4/136   15/136  14/136  1/136

What is:

a.) P(X = Y)
b.) P(X + Y = 5)
c.) P(1 < X <= 3, 1 < Y <= 3)
d.) p((X,Y) $$\in$$ {1,4} x {1,4})

2. Relevant equations

3. The attempt at a solution

To be honest, I'm not exactly sure what exactly these kinds of operators mean (e.g. what does it mean to say P(X = Y). But here's what I have so far:

P(a = 1) = P(a = 2) = P(a = 3) = P(a = 4) = P(b = 1) = P(b = 2) = P(b = 3) = P(b = 4) = 34/136

a.) P(X = Y) is [P(a = 1, b = 1) + ... + P(a = 4, b =4)] = 16/136 + 10/136 + 7/136 + 1/136

b.) P(X + Y = 5) = [P(a = 1, b = 4) + P(a = 4, b = 1)] = 13/136 + 4/136

c.) P(1 < X <= 3, 1 < Y <= 3) = c.) P(1 < a <= 3, 1 < b <= 3) =
10/136 + 11/136 + 6/136 + 7/136

d.) p((X,Y) $$\in$$ {1,4} x {1,4}) = ? I have no idea what this syntax means

2. Oct 18, 2009

### Staff: Mentor

I don't know what you mean by the equations above. The probabilities in the table are joint probabilities.
Looks good (above).
What other ways can X + Y add to 5? You're showing just two of them. (above).
Looks good (above).
It's asking for P((X, Y)) where X can be 1, 2, 3, or 4 and Y can be 1, 2, 3, 4.

3. Oct 18, 2009

### whs

Part 'd' means the probability that the pair X,Y will be in the Cartesian product of {1,4} x {1,4}. Which means, P(X=1, Y=1) or P(X=2, Y=1) or ... and so on, for all pairs in that product.

Seeing your table, it looks like that would cover all the possible values in that table, so it would be 1.