Geometric Distribution: Definition, Equations & Examples

Instructor: Damien Howard

Damien has a master's degree in physics and has taught physics lab to college students.

Discover what the geometric distribution is and the types of probability problems it's used to solve. Then, solidify everything you've learned by working through a couple example problems.

Probability Distributions

If you want to know the probability that an outcome of an event will occur, what you're looking for is the likelihood that this outcome happens over all other possible outcomes. The probability of an outcome occurring could be a simple binary 50/50 choice, like whether a tossed coin will land heads or tails up, or it could be much more complicated.

No matter how complicated, the total sum for all possible probabilities of an event always comes out to 1. The manner in which these probabilities all add up to 1 is called their probability distribution.

There are many types of probability distributions, each one used for specific situations. In this lesson we're going to learn about the geometric distribution.

Geometric Distribution

To understand what the geometric distribution is used for, we have to first start with something called a Bernoulli trial. A Bernoulli trial is any experiment that has exactly two possible results, such as the example of a coin being tossed.

In a Bernoulli trial, we label one of the two possible results as success and the other as failure. A geometric distribution is the probability distribution for the number of identical and independent Bernoulli trials that are done until the first success occurs.

In order to use the geometric distribution to find probabilities, we use the following formulas, where each one corresponds to the given specific circumstance.

In the table, p is the probability of success of a single Bernoulli trial, q is the probability of failure of a single Bernoulli trial, and Y is a discrete random variable that can take on any of the possible values given by y.

The P(Y > y) and P(Y > y) probabilities can't be solved directly with a sum like the formulas with less than signs because the range of y extends infinitely. Summing this directly would mean you need to sum from some value to infinity. We can get around this because we know the sum of all probabilities P(Y ≥ 1) must equal 1, and P(Y ≥ 1) can be split into two halves added together that sum over the full range of y.

Mean and Variance

Each probability distribution has its own unique formula for mean and variance of the random variable Y. The mean or expected value of Y tells us the weighted average of all potential values for Y. For a geometric distribution mean (E(Y) or μ) is given by the following formula.

The variance of Y is defined as a measure of spread of the distribution of Y. The variance (V(Y) or σ2) for a geometric random variable is written as follows.

Example Problems

In order to cement everything we've gone over in our heads, let's work through an example problem together.

A coin has been weighted so that it has a 0.9 chance of landing on heads when flipped. What is the probability that the first time the coin lands on heads is after the 3rd flip?

In this problem, we're asked to find the probability that the first success happens after the 3rd Bernoulli trial. Using our chart from earlier, we can see that we want to use the P(Y > y) form of the formula with 3 substituted in for y.

Next, we need the probability of failure of a single Bernoulli trial (q). This can be found because the total probability of a Bernoulli trial must equal 1, and that probability only has two parts; the probability of success (p) and the probability of failure (q).

Using the probability table from earlier to find P(Y ≤ 3), we now have everything needed to solve this problem.

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