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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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?
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?
I understand these three cases but I've no idea about the fourth option. What does that mean? These tree cases mean 10 different ways. What's the fourth?Quote:
Originally Posted by Jameson
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?
Am I right when I say that the answer is 10/(4*4*4)=10/64Quote:
Originally Posted by Jameson
Looks right to me. Hopefully one of the staff members doesn't disagree... :)
I think it makes sense and is correct.
Hello,
Sorry for the autopromootion of my work lol!
There is a formula for the number of combinations for x+y+z=3 and the solution is(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