# Thread: [SOLVED] (Simple?) Discrete joint function help

1. ## [SOLVED] (Simple?) Discrete joint function help

Hello, I have small problem with this problem, here it is

Question :
We pick 3 persons at random from a group of 13 in which exactly 5 have blue eyes.

Let Xi = { 1 if the i ème person has blue eyes
0 if not

What is the joint mass function of X1 and X2?

My actual attempt to a solution
I know that when X1=0 that P(X1=0)=8/13, P(X2=0)=7/12 and P(X2=1)=5/12

and when X1=1 that P(X1=1)=5/13, P(X2=0)=8/12 and P(X2=1)=4/12

I also know that the joint mass function is p(x1, x2) = P(X1 = x1 and X2 = x2)

My question is : so do I write P(X1=0 and X2=0) = P(X1=0) * P(X2=0) = 8/13*7/12
or do I simply write P(X1=0 and X2=0) = (8/13,7/12) or something else???

Thanks a lot.

2. Incorrect
P(X2=0)=7/12 is false
P(X2=0|X1=0)=7/12
while P(X2=0)=8/13.
It doesn't matter when you select.
If you don't have any info on the first selection the probability is the same.

The probability that the first card from a deck is the ace of spades is 1/52.
The probability that the tenth card is the aces of spades is also 1/52.
If you give me some info on the first nine cards, then the probability changes.

So P(X_1=0, X_2=0)=P(X_2=0|X_1=0)P(X_1=0)= (7/12)(8/13)
In general
P(X_1=a, X_2=b)=P(X_2=b|X_1=a)P(X_1=a).

3. Likewise if we have an urn with 3 red and 4 blue balls and we select without replacement,
the probability the second ball is red is STILL 3/7.
It's the same probability as the first selection, since we have no info on the first pick.
P(second ball is red)=P(first is red and second is red)+P(first is blue and second is red)
Using conditional probabilities we get..
=P(second is red|first is red)P(first is red)+
P(second is red|first is blue)P(first is blue)
=(2/6)(3/7)+(3/6)(4/7)
=3/7.

4. Ok excellent so,

P(X_1=0)=8/13 and P(X_1=1)=5/13

P(X_2=0|X_1=0)=7/12 P(X_2=1|X_1=0)=5/12
P(X_2=0|X_1=1)=8/12 P(X_2=1|X_1=1)=4/12

so for my join mass function of X_1 and X_2 I have:
P(X_1=0, X_2=0)=P(X_2=0|X_1=0)P(X_1=0) = 7/12 * 8/13 = 56/156 = 28/78 = 14/39
P(X_1=0, X_2=1)= 5/12 * 8/13 = 40/156 = 20/78 = 10/39
P(X_1=1, X_2=1)= 4/12 * 5/13 = 20/156 = 10/78 = 5/39
P(X_1=1, X_2=0)= 8/12 * 5/13 = 40/156 = 10/39

Right?

Thanks.

5. I'm too tired to check if you multiplied correctly.
But, yes these fractions are correct.
Make sure all 4 probabilities sum to one.

You were confusing conditional probabilities with marginals.
I need to go to bed.
I teach two prob and stat classes in the morning.

6. Thanks a ton