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July 18th, 2019, 06:03 AM   #1
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Independent discrete random variables probability

Independent random variables X, Y, Z take only integer values: X - from 0 to 7, Y - from 0 to 10, Z - from 0 to 13. Find the probability P (X + Y + Z = 4) if it is known that the possible values of X, Y, and Z are equiprobable. The solution is attached below; however, I don't understand where the number of combinations of 6 objects of 2 comes from. Could you please help me with that?
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Last edited by skipjack; July 19th, 2019 at 03:02 AM.
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July 18th, 2019, 12:16 PM   #2
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Any partition of 4 with non-negative integers will limit the values of X,Y,Z to the range of [0,4]

So for success, i.e. that X+Y+Z=4, we need only consider that range for each variable.

Given that, the problem is equivalent to taking 4 balls and arranging them into the 3 bins
labeled X, Y, and Z.

This is a stars and bars problem and the number of arrangements is known to be

$\dbinom{4+3-1}{3-1} = \dbinom{6}{2}$

That's where the $C_6^2$ comes from.
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Last edited by romsek; July 18th, 2019 at 01:04 PM.
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July 19th, 2019, 05:31 AM   #3
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Alright, how then to solve the problem with the same conditions but find the P(X ≠ Y, Y ≠ Z, Z≠X)? Cause in that case method of stars and bars doesn't work as far as i understand
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July 19th, 2019, 06:36 AM   #4
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Without a formula, the amount of possibilities is small enough to list them:

0, 0, 4
0, 1, 3
0, 2, 2
0, 3, 1
0, 4, 0
1, 0, 3
1, 1, 2
1, 2, 1
1, 3, 0
2, 0, 2
2, 1, 1
2, 2, 0
3, 0, 1
3, 1, 0
4, 0, 0
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