Geometric distribution

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-1p 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 σ² = (1 - p)/p² 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-1p.

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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) = 0.985⁹⁹0.015 + 0.985¹⁰⁰0.015 + 0.985¹⁰¹0.015 + ...

P(X ≥ 100) = 0.985⁹⁸(0.015)(0.985 + 0.985² + 0.985³ + ...)

* Recall that z + z² + z³ + ... is the geometric series, and it approximates z/(1-z). Thus:

P(X ≥ 100) = 0.985⁹⁸(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!