# Geometric Distribution

**Episode #7 of the course Theory of probability by Polina Durneva**

Good morning!

Today, we will have a discussion on another discrete probability distribution: geometric distribution. Geometric distribution is based on Bernoulli distribution, as it demonstrates how many failures you get until your first success.

**What Is Geometric Distribution?**

In geometric distribution, you have several trials. The outcome for each trial is either success (normally denoted as 1) or failure (denoted as 0). Each trial, therefore, follows Bernoulli distribution. You might think that geometric distribution resembles binomial distribution too. However, binomial distribution represents the number of successes for a given set of trials. This number can be anything: from 0 to the total number of trials. In geometric distribution, you have only one success: After getting this one success, you don’t do any more trials.

For example, if you want to know the probability of getting 6 on your 3rd trial of rolling a die, you will use geometric distribution. However, if you want to know the probability of getting two 6s while rolling a die 3 times, you will use binomial distribution. Hopefully, this difference is clear enough. Before we jump into formulas, expected values, and variance, let’s discuss the main assumptions for geometric distribution:

1. Each trial is either success or failure.

2. All trials are independent.

3. The probability of success and failure is consistent from one trial to another.

**Formula for Geometric Distribution**

The formula for geometric distribution takes the following form: P(x) = p * (1 – p)^{x-1}, where x is the number of trials. Let’s proceed to an example to better the above-mentioned formula. Assume you are rolling a die 3 times and want to know the probability of getting a 6 on your last trial. In this case, you want to know the probability of success after getting two failures. The probability of success is 1/6, and the probability of failure is 1 – 1/6 = 5/6. Applying the formula, you will get: (1/6) * (5/6)^{2} = 25/216 = 11.6%.

**Expected Value of Geometric Distribution**

The expected value, or the average value, of geometric distribution can be calculated by dividing 1 by the probability of success (1/p). In our previous example, the probability of success (or getting a 6) is 1/6. The expected value is then 6, meaning that it takes, on average, 6 trials to get one success.

**Variance of Geometric Distribution**

Variance of geometric distribution is a bit more complicated and requires time to be proved. In this course, we’ll skip the proof and go straight to the formula: p/(1 – p)^{2}, where p is the probability of success. Using our previous example, we can find variance. It would be: (1/6) / (1 – 1/6)^{2} = 6/25 = 0.24, which shows the spread of the given data.

That’s it for our lesson today! Tomorrow, we will have a discussion on our final discrete probability distribution in this course (future courses might cover more discrete probability distributions): negative binomial distribution, which resembles geometric and binomial distributions to an extent.

Can’t wait to see you in the morning!

Have a great day,

Polina

**Recommended book**

*Probability Theory: A Concise Course* by Y.A. Rozanov

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