My Math Forum simple probability

 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 $\frac{1}{2^n}$. 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 $\leq$ 3) = $\frac{1}{8} + \frac{1}{4} + \frac{1}{2}$ = $\frac{7}{8}$ So P(X > 3) = 1 - 7/8 = 1/8 So P(X > 3 AND Y $\leq$ 3) = P(X>3)*P(Y $\leq$ 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 $\leq$ 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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