In a series of trials, if you assume that the probability of either success or failure of each trial is the same, geometric distribution gives the probability of achieving success after N number of failures. The distribution is essentially a set of probabilities that presents the chance of success after zero failures, one failure, two failures and so on.

## Geometric Distribution Formula

In probability and statistics, geometric distribution defines the probability that first success occurs after k number of trials. If p is the probability of success or failure of each trial, then the probability that success occurs on the \(k^{th}\) trial is given by the formula

\(Pr (X = k) = (1-p)^{k-1}p\)

## Geometric Distribution Examples

Consider a couple who are planning to have a child and they will continue to babies until it is a girl. What is the probability that they have zero boys, one boy and two boys and so on until a girl is born?

A person is seeking new employment that is both challenging and fulfilling. What is the probability that he will quit zero times, one time, two times so on until he finds his ideal job?

A pharmaceutical company is designing a new drug to treat a certain disease that will have minimal side effects. What is the probability that zero drugs fail the test, one drug fails the test, two drugs fail the test and so on until they have designed the ideal drug?

## Assumptions for Geometric Distribution to Be an Ideal Model

- The model is appropriate for a phenomenon that has a series of trials
- The model is appropriate if each trial has only two possible outcomes – either success or failure
- The model is appropriate only if the probability of success is the same for each trial

## Probability of Outcomes

Let us consider one of the situations where geometric distribution can be applied. Let us say a person is throwing a dice and will stop once he gets 5. Since there are 6 possible outcomes, the probability of success p = \(\frac{1}{6}\) = 0.17. Therefore the probability of failure

q = 1 – p = 1 – 0.17 = .83

Let us calculate the probability of first three trials.

The person gets number 5 for the first time. The number of failures before the first success is zero. Therefore X = 0, k = 1

Substituting the values of X, k, p and q in geometric distribution, we have

\(Pr(X=0)= (0.83^{0})\times 0.17\) = 0.17

The person gets number 5 for the second time. The number of failures before the first success is 1. Therefore X = 1 and k = 2.

Substituting the values of X, k, p and q in geometric distribution, we have

\(Pr(X=1)= (0.83^{2-1})\times 0.17\) = 0.83 x 0.17 = 0.14

This way we can construct a series of geometric distribution for a series of trials.

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