Discrete Maximum

[booyaolian]

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

Compute P[max(X,Y)=6].

[atomic]

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.

[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

[booyaolian]

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....

[daaaave]

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.

[booyaolian]

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...