• Welcome! The new FreeMathHelp.com forum is live. We've moved from VB4 to Xenforo 2.1 as our underlying software. Hopefully you find the upgrade to be a positive change. Please feel free to reach out as issues arise -- things will be a little different, and minor issues will no doubt crop up.

Joint PMF (calculating probabilities)

thumper88

New member
Joined
Apr 3, 2012
Messages
1
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=0y=1y=2y=3y=4P[SUB]X[/SUB](x)
x=0
x=10
x=200
x=3000
x=40000
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
Joined
Jan 29, 2005
Messages
7,688
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.
 
Top