The Geometric distribution is a probability distribution that is used to model the probability of experiencing a certain amount of failures before experiencing the first success in a series of Bernoulli trials.
A Bernoulli trial is an experiment with only two possible outcomes – “success” or “failure” – and the probability of success is the same each time the experiment is conducted.
An example of a Bernoulli trial is a coin flip. The coin can only land on two sides (we could call heads a “success” and tails a “failure”) and the probability of success on each flip is 0.5, assuming the coin is fair.
If a random variable X follows a geometric distribution, then the probability of experiencing k failures before experiencing the first success can be found by the following formula:
P(X=k) = (1-p)kp
- k: number of failures before first success
- p: probability of success on each trial
In this article we share 5 examples of how the Geometric distribution is used in the real world.
Example 1: Coin Tosses
Suppose we want to know how many times we’ll have to flip a fair coin until it lands on heads.
We can use the following formulas to determine the probability of experiencing 0, 1, 2, 3 failures, etc. before the coin lands on heads:
Note: The coin can experience 0 “failures” if it lands on heads on the first flip.
P(X=0) = (1-.5)0(.5) = 0.5
P(X=1) = (1-.5)1(.5) = 0.25
P(X=2) = (1-.5)2(.5) = 0.125
P(X=3) = (1-.5)3(.5) = 0.0625
Example 2: Supporters of a Law
Suppose a researcher is waiting outside of a library to ask people if they support a certain law. The probability that a given person supports the law is p = 0.2.
We can use the following formulas to determine the probability of interviewing 0, 1, 2 people, etc. before the researcher speaks with someone who supports the law:
P(X=0) = (1-.2)0(.2) = 0.2
P(X=1) = (1-.2)1(.2) = 0.16
P(X=2) = (1-.2)2(.2) = 0.128
Example 3: Number of Defects
Suppose it’s known that 5% of all widgets on an assembly line are defective.
We can use the following formulas to determine the probability of inspecting 0, 1, 2 widgets, etc. before an inspector comes across a defective widget:
P(X=0) = (1-.05)0(.05) = 0.05
P(X=1) = (1-.05)1(.05) = 0.0475
P(X=2) = (1-.05)2(.05) = 0.04512
Example 4: Number of Bankruptcies
Suppose it’s known that 4% of individuals who visit a certain bank are visiting to file bankruptcy. Suppose a banker wants to know the probability that he will meet with less than 10 people before encountering someone who is filing for bankruptcy.
We can use the Geometric Distribution Calculator with p = 0.04 and x = 10 to find that the probability that he meets with less than 10 people before encountering someone who is failing for bankruptcy is 0.33517.
Example 5: Number of Network Failures
Suppose it’s known that the probability that a a certain company experiences a network failure in a given week is 10%. Suppose the CEO of the company would like to know the probability that the company can go 5 weeks or longer without experiencing a network failure.
We can use the Geometric Distribution Calculator with p = 0.10 and x = 5 to find that the probability that the company lasts 5 weeks or longer without a failure is 0.59049.