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April 7th, 2010, 09:41 PM  #1 
Senior Member Joined: Dec 2009 Posts: 150 Thanks: 0  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. 
April 8th, 2010, 09:23 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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.

April 8th, 2010, 11:36 AM  #3 
Senior Member Joined: Dec 2009 Posts: 150 Thanks: 0  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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