Geometric Probability Distribution

In this lesson, we cover the geometric probability distribution, which is a special case of the negative binomial distribution.

A Geometric Experiment

A geometric experiment is a statistical experiment that has the following properties:

• The experiment consists of x repeated trials.
• Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
• The probability of success, denoted by P, is the same on every trial.
• The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
• The experiment continues until the first success is observed.

An example of a geometric distribution would be tossing a coin until it lands on heads. We might ask: What is the probability that the first head occurs on the third flip? That probability is referred to as a geometric probability.

Geometric Probability

The formula for geometric probability is given below.

Simple formulas also exist to compute the cumulative probability that (a) at most x trials are required to achieve the first success in a geometric experiment or (b) at least x trials are required to achieve the first success. Here are those formulas:

• At most x trials: P(X ≤ x) = 1 - (1 - P)^x
• At least x trials: P(X ≥ x) = (1 - P)^x-1

Geometric Probability Distribution

A geometric random variable is the number X of repeated trials to produce the first success in a geometric experiment. The probability distribution of a geometric random variable is called a geometric probability distribution

Suppose we flip a coin repeatedly and count the number of trials until the coin has landed for the first time on heads. The geometric random variable is the number of coin flips required to land for the first time on heads. In this example, the number of coin flips can take on any integer value between 1 and plus infinity. The geometric probability distribution for this example is presented below.

Number of coin flips	Probability
1	0.5
2	0.25
3	0.125
4	0.0625
5	0.03125
6 or more	0.03125

Geometric Distribution: Mean, Variance, and Mode

If we define the mean of the geometric distribution as the average number of trials required to produce the first success, then the mean and variance of the geometric probability distribution are equal to:

μ = 1 / P

σ² = (1 - P) / P²

where μ is the mean of the geometric distribution, σ² is the variance of the geometric distribution, and P is the probability of a success on any given trial.

The mode of a geometric distribution is always 1, regardless of probability. This is because the most likely outcome is that the first success happens immediately on the first trial.

Test Your Understanding

Problem 1: Geometric Probability

Bob is a high school basketball player. He is a 70% free throw shooter. That means his probability of making a free throw is 0.70. What is the probability that Bob makes his first free throw on his fifth shot?

Solution: This is an example of a geometric experiment. Therefore, this problem can be solved using the geometric formula.

The probability of success (P) is 0.70, and the number of trials (x) is 5. We enter these values into the geometric probability formula.

P(X=5) = P * (1 - P)^{x - 1}
P(X=5) = 0.7 * 0.3⁴ = 0.00567

Problem 2: At Least

A factory produces defective light bulbs with a probability of 0.02. What is the probability that it takes more than 15 tests to find a defective bulb?

Solution: We want to know the probability that it takes at least 16 tests to find a defective bulb, when the probability of finding a defective bulb is 0.02. Therefore,

P(X ≥ 16) = = (1 - P)^x-1
P(X ≥ 16) = = (0.98)¹⁵ = 0.7386

Problem 3: At Most

A salesperson has a 10% chance of making a sale per call.What is the probability it takes at most 7 sales calls to make the first sale?

Solution: We want to know the probability that it takes 7 or fewer sales calls to make a sale, when the probability of making a sale is 0.1. Therefore,

P(X ≤ 7) = 1 - (1 - P)^x
P(X ≤ 7) = 1 - (0.9)⁷ = 0.522

Problem 4

The chance of winning a small prize in a lottery is 0.05 per ticket.

• What is the mean number of tickets needed to win the first prize?
• Which ticket is most likely to be the first winning ticket?

Solution: The mean number of tickets needed to win the first prize is:

μ = 1 / P = 1 / 0.05 = 20 tickets

The mode of a geometric distribution is always 1, regardless of probability. The first success is more likely to occur on the first trial than on any other trial. Therefore, the first ticket purchased is more likely than any other ticket to be the first winning ticket.