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#1
09-26-2008, 02:04 AM
 booyaolian Member Join Date: Feb 2007 Posts: 125
Discrete Maximum

Let X and Y be independent and each uniformly distributed on {1,2,3,...,15}.

Compute P[max(X,Y)=6].
#2
09-26-2008, 02:32 AM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

If max{X,Y} = 6, then there are exactly three mutually disjoint cases:

1. X = 6, Y < 6
2. X < 6, Y = 6
3. X = Y = 6.

What is the probability of the first case? Since X and Y are independent,

Pr[X = 6 and Y < 6] = Pr[X = 6] Pr[Y < 6].

The rest is straightforward.
#3
09-26-2008, 03:21 AM
 procrastinator Member Join Date: Sep 2006 Location: Cali College: third time is a charm Favorite beer: Dogfish Posts: 1,906

It is the probability they are both at most 6 minus the probability they are both at most 5.

(6/15)^2-(5/15)^2
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#4
09-26-2008, 01:18 PM
 booyaolian Member Join Date: Feb 2007 Posts: 125

Quote:
 Originally Posted by procrastinator It is the probability they are both at most 6 minus the probability they are both at most 5. (6/15)^2-(5/15)^2
I think this one is right....
#5
09-26-2008, 01:41 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,486

Quote:
 Originally Posted by booyaolian I think this one is right....
Atomic's answer is also right, although procrastinator's method is the one that I usually recommend on this type of problem.
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#6
09-26-2008, 05:00 PM
 booyaolian Member Join Date: Feb 2007 Posts: 125

Quote:
 Originally Posted by daaaave Atomic's answer is also right, although procrastinator's method is the one that I usually recommend on this type of problem.
Please ignore the last retarded response. My girlfriend posted it...

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