What's the probability that x+y+z=3

if the values of x, y and z can be 0, 1, 2, 3?

It seems to me that all possible ways to make a sum are 3*3*3=27. 27 ways to sum up numbers. How to figure out the sums that x+y+z=3?

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- Sep 23rd 2008, 10:51 AMtotalnewbieProbabilty that x+y+z=3
What's the probability that x+y+z=3

if the values of x, y and z can be 0, 1, 2, 3?

It seems to me that all possible ways to make a sum are 3*3*3=27. 27 ways to sum up numbers. How to figure out the sums that x+y+z=3? - Sep 23rd 2008, 10:54 AMJameson
You have one case where all three equal 1. Then the only other ways to make 3 are by 3+0+0 and 2+1+0. So figure out all the arrangements of those two sums.

EDIT: Also, you have FOUR options to choose from for every term in the sum, not three. See what I mean? - Sep 23rd 2008, 11:05 AMtotalnewbie
- Sep 23rd 2008, 11:18 AMJameson
Sorry, I wasn't clear enough. I meant that when figuring out all the possible ways to make a sum, you calculated it assuming x,y,z had 3 values to choose from, when it seems there are 4, no?

- Sep 23rd 2008, 11:22 AMtotalnewbie
- Sep 23rd 2008, 11:32 AMJameson
Looks right to me. Hopefully one of the staff members doesn't disagree... :)

I think it makes sense and is correct. - Sep 23rd 2008, 11:39 AMMoo
Hello,

Sorry for the autopro*moo*tion of my work lol!

There is a formula for the number of combinations for x+y+z=3 and the solution is $\displaystyle {5 \choose 3}=10$ (See here for a kind of explanation : http://www.mathhelpforum.com/math-he...ial-proof.html )

So there are 10 possibilities that are interesting for us (which is the same result as what you got) and there is a total of 4^3=64 ways to combine x,y and z.

Thus the answer is 10/64