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July 18th, 2019, 06:03 AM  #1 
Newbie Joined: Jun 2019 From: London Posts: 13 Thanks: 0  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?
Last edited by skipjack; July 19th, 2019 at 03:02 AM. 
July 18th, 2019, 12:16 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1475 
Any partition of 4 with nonnegative 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 $\dbinom{4+31}{31} = \dbinom{6}{2}$ That's where the $C_6^2$ comes from. Last edited by romsek; July 18th, 2019 at 01:04 PM. 
July 19th, 2019, 05:31 AM  #3 
Newbie Joined: Jun 2019 From: London Posts: 13 Thanks: 0 
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

July 19th, 2019, 06:36 AM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88 
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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discrete, independent, probability, random, variables 
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