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April 7th, 2010, 09:41 PM   #1
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simple probability

Suppose X is a discrete random variable on the set of positive integers such that for each positive integer n, the probability that X = n is . If Y is another random variable with the same probability distribution and X and Y are independent, what is the probability that the value of at least on of the variables X and Y is greater than 3?

Is the following correct?
P(X 3) = =
So P(X > 3) = 1 - 7/8 = 1/8

So P(X > 3 AND Y 3) = P(X>3)*P(Y 3) = (1/*(7/ = 7/64.
I believe you can separate the probability like I just did because of the independence, no?

Then P(Y > 3 AND X 3) = 7/64 too because it's the same thing really when they're independent.

The last case is both of them greater than 3, which is
P(X > 3 AND Y > 3) = P(X > 3)*P(Y > 3) = (1/(1/ = 1/64

Since any one of these possibilities satisfies the requirement, we simply add their probabilities (OR -> + , AND -> *) to get
7/64 + 7/64 + 1/64 = 15/64

please tell me if I made any mistakes, I know this is the right answer, but I want my thought process to be right too.
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April 8th, 2010, 09:23 AM   #2
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Re: simple probability

That's fine. Somewhat simpler: 7/8 chance of being at most 3, so 49/64 chance of both being at most 3, so 1 - 49/64 = 15/64 of at least one being greater than 3.
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April 8th, 2010, 11:36 AM   #3
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Re: simple probability

Tremendously helpful. This was a GRE math subject test question, and of course speed is very important. Thanks for confirming my tedious reasoning to boot.
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