# Joint PMF (calculating probabilities)

#### thumper88

##### New member
So I have a problem that says four fair dice are rolled. So that tells me 6[SUP]4[/SUP] sample space.
X is the number of fives observed and Y is the number of sixes observed.
I created a table
 P[SUB]X,Y[/SUB](x,y) y=0 y=1 y=2 y=3 y=4 P[SUB]X[/SUB](x) x=0 x=1 0 x=2 0 0 x=3 0 0 0 x=4 0 0 0 0 P[SUB]Y[/SUB](y)

I understand the method of calculating the PMF, I'm having counting issues, calculating the probability for each part. I believe the zeros are placed correctly? If 4 fives are observed you can't have more than 0 sixes observed and vice versa.
For example, how do I calculate the probability of 2 fives observed and 2 sixes observed? Please help, thanks.

#### pka

##### Elite Member
So I have a problem that says four fair dice are rolled. So that tells me 6[SUP]4[/SUP] sample space.
X is the number of fives observed and Y is the number of sixes observed.
LaTeX is down so a fancy answer is out of the question .
Let #(x=m,y=n) mean the number of 4-tuples in which there are m fives and n sixes.
#(x=0,y=0)=4[SUP]4 [/SUP],

#(x=1,y=2)=(4!/2)(4)=2(4!).

#(x=2,y=2)=6

You try the rest.