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### Probability Using Geometric Random Variables

```Date: 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?

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

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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