Discrete

distribution used to find the

probability of success after a certain number of

trials, in which "success" is defined as occurance of a certain

event. If

*X* is a

random variable representing the number of

independent trials until success, and

*p* is the fixed probability of success for each trial, then we say

*X* ~

Geometric(

*p*). The

probability density function is given as follows:

*P*(*X* = *x*) = ƒ(*x*) = (1 - *p*)^{x-1}*p* for *x* = 1, 2, ...

P(*X* = *x*) = *ƒ*(*x*) = 0 otherwise

With the geometric distribution, the mean and population variance are given by *μ* = 1/*p* and *σ*^{2} = (1 - *p*)/*p*^{2} respectively.

The origin of the pdf is relatively easy to understand. The probability of success on the *x*^{TH} trial is equal to (*x* - 1) failures (each with probability of (1 - *p*)) and one success (with probability *p*). So multiply the two probabilities and you get (1 - *p*)^{x-1}*p*.

__Example__: **Suppose the probability that a randomly selected E2 writeup is deleted by dannye is 0.015. Find the probability that none of your next 100 writeups have a date with dannye's "kill" button.**

Let *X* be a random variable for the number of writeups you create before dannye kills one. Assume *X* ~ Geometric(*0.015*)

*P*(*X* ≥ 100) = *P*(*X* = 100) + *P*(*X* = 101) + *P*(*X* = 102) + ...

*P*(*X* ≥ 100) = 0.985^{99}0.015 + 0.985^{100}0.015 + 0.985^{101}0.015 + ...

*P*(*X* ≥ 100) = 0.985^{98}(0.015)(0.985 + 0.985^{2} + 0.985^{3} + ...)

* Recall that z + z^{2} + z^{3} + ... is the geometric series, and it approximates z/(1-z). Thus:

*P*(*X* ≥ 100) = 0.985^{98}(0.015)(0.985/0.015) = 0.224

**So, there's a 22.4% chance that dannye won't kill one of your next 100 writeups!**