Probability Using Geometric Random VariablesDate: 01/15/2004 at 00:47:40 From: Jamie Subject: statistics/probability Let X be a geometric random variable with parameter p1 and let Y be a geometric random variable with parameter p2. Find the probability that X is less than or equal to Y. I'm not too sure where to even begin with this problem or what to do. I know what the geometric distribution is, and I thought of possibly conditioning on X to solve this problem. Date: 01/15/2004 at 09:30:06 From: Doctor Mitteldorf Subject: Re: statistics/probability Hi Jamie, Here's how you might start: Think of the geometric random variable as the probability of requiring X trials to turn up your first success, if the probability for success on each trial is p. So let's say you are flipping a biased coin so the probability of heads is p1 and I am flipping another biased coin with a probability of heads equal to p2. What is the probability that you will flip your first heads either before my first heads or at the same time? Answer: Let's use q1 to mean 1 - p1 and q2 = 1 - p2. Your very first flip, you might get heads with probability p1, and then you win. Otherwise you might get heads on your second try (q1*p1), and you win if I get tails on my first try (q2). So far we have p1 + q1*p1*q2 Continuing, the probability of your winning on the third flip is q1*q1*p1*q2*q2. Write down the fourth and fifth terms yourself, and keep going until you see a pattern. Once you have the pattern, see if it leads you to an idea for finding the infinite sum of all the terms. Please write back if a further hint would be useful, or let me know when you've got an answer! - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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