A **geometric distribution** tells us how many trials (*k*) are required until we obtain the first “success.”

The probability that we will obtain the first success on the k^{th} trial can be found using the formula:

P(obtain first success on k^{th} trial) = (1-p)^{k-1} * p

where *p *is the probability of success in a given trial.

The geometric distribution has the following properties:

- The mean of the distribution is
**μ**= 1/p - The variance of the distribution is
**σ**= (1-p) / p^{2}^{2} - The standard deviation of the distribution is
**σ**= √σ^{2}

This tutorial explains how to solve problems using the geometric distribution in R.

*You can find a complete introduction to the geometric distribution here.*

**Geometric Distribution in R: Syntax**

The two built-in functions in R we’ll use to answer questions using the geometric distribution are:

**dgeom(x, prob) **– calculates the probability mass function (pmf) for the geometric distribution where *x *is the number of failures before the first success and *prob *is the probability of success on each trial.

**pgeom(q, prob) **– calculates the cumulative distribution function (cdf) for the geometric distribution where *q *is the number of failures before the first success and *prob *is the probability of success on each trial.

*Find the full R documentation for the geometric distribution here.*

**Solving Problems Using the Geometric Distribution in R**

**Example 1: ***A restaurant gives a raffle ticket to each customer that walks in the door. If a customer receives a winning raffle ticket, they get a free dinner. The probability that a given raffle ticket is a winner is 5%. What is the probability that the 10 ^{th} customer that walks in the door is the first winner?*

**Solution:** We want to know the probability that there will be nine “failures” (people not receiving a winning ticket) before the first “success” (person receives a winning ticket). Since we know the probability of “success” for each person is 0.05, we will use the following syntax in R to find the solution:

dgeom(9, 0.05)

## [1] 0.0315124704862305

The probability that the 10th customer that walks in the door is the first winner is **0.0315**.

**Example 2: ***Jessica **makes 70% of her free-throw attempts. What is the probability that she makes her first free-throw within her first two attempts?*

**Solution:** We want to know the probability that it will take two attempts at most for Jessica to make her first free-throw. Since we know the probability of “success” for each attempt is 0.7, we will use the following syntax in R to find the solution:

#calculate the cumulative probability of experiencing either 0 or 1 failures #before the first "success" occurs, given that probability of success for each #attempt is 0.7.pgeom(1, 0.7)

## [1] 0.91

Thus, the probability that Jessica makes her first free-throw within her first two attempts is **0.91**.

**Example 3: Mike makes 80% of his field goal attempts. What is the probability that he needs more than two attempts to make his first field goal?**

**Solution:** To answer this question, we can use the formula 1 – (probability that he needs two or less attempts to make first field goal). This is given by:

1-pgeom(1, .8)

## [1] 0.04

The probability that Mike needs more than two attempts to make his first field goal is **0.04**.