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The geometric distribution is a probability distribution that models the number of successes in a sequence of independent trials. The geometric distribution is often used in real-world situations, such as the number of defective items in a shipment, the number of phone calls before a customer service agent answers, the number of times a light bulb needs to be switched before it turns on, the number of clicks before a website visitor purchases an item, and the number of times a patient needs to be treated before a medical condition is cured.
The 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 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
where:
- 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.
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 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 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.