Study GuidesExam QuestionsPast Papers
AP®StatisticsCollege BoardStudy GuidesUnit 4: Probability, Random Variables & Probability DistributionsBinomial & Geometric DistributionsProbabilities for Geometric Distributions

Probabilities for Geometric Distributions (College Board AP® Statistics)

Study Guide

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Probabilities for geometric distributions

How do I calculate the probability of a single outcome using a geometric distribution?

  • Let X be a geometric random variable with the parameter p

    • X tilde G e o open parentheses p close parentheses

  • The formula for P open parentheses X equals x close parentheses, where x equals 1 comma space 2 comma space 3 comma space... comma space is

    • P open parentheses X equals x close parentheses equals open parentheses 1 minus p close parentheses to the power of x minus 1 end exponent p

      • This is given in the exam

    • e.g. if X tilde G e o open parentheses 0.7 close parentheses then P open parentheses X equals 5 close parentheses equals open parentheses 1 minus 0.7 close parentheses to the power of 5 minus 1 end exponent open parentheses 0.7 close parentheses

      • this can be written as P open parentheses X equals 5 close parentheses equals open parentheses 0.3 close parentheses to the power of 4 open parentheses 0.7 close parentheses

How do I calculate the probability of an event using a geometric distribution?

  • To calculate the probability of an event

    • find the probabilities of each outcome in the event

    • add the probabilities together

      • e.g. P open parentheses 2 less or equal than X less than 6 close parentheses equals P open parentheses 2 close parentheses plus P open parentheses 3 close parentheses plus P open parentheses 4 close parentheses plus P open parentheses 5 close parentheses

  • To calculate the probability of an event using the complement of the event

    • find the probabilities of the outcomes not in the event

    • add the probabilities together

    • subtract from 1

      • e.g. P open parentheses X greater than 3 close parentheses equals 1 minus open parentheses P open parentheses 0 close parentheses plus P open parentheses 1 close parentheses plus P open parentheses 2 close parentheses plus P open parentheses 3 close parentheses close parentheses

  • If the first success occurs no earlier than trial x, then the first x minus 1 trials must have been failures

    • You can use the formula P open parentheses X greater or equal than x close parentheses equals open parentheses 1 minus p close parentheses to the power of x minus 1 end exponent where x equals 1 comma space 2 comma space 3 comma space...

      • This formula is not given in the exam

  • If the first success occurs no later than trial x, then the first x trials must not have all been failures

    • You can use the formula P open parentheses X less or equal than x close parentheses equals 1 minus open parentheses 1 minus p close parentheses to the power of x where x equals 1 comma space 2 comma space 3 comma space...

      • This formula is not given in the exam

What is the memoryless property of the geometric distribution?

  • The memoryless property means that if the first x trials end in failure, then they can be ignored when calculating the probability of the first success after these trials

    • e.g. given that the first 10 rolls of a fair dice did not land on a six, the probability that the 11th roll will land on a six is equal to 1 over 6

      • This is equal to the probability that the first dice roll lands on a six

  • Mathematically, this property can be expressed as P open parentheses X equals x plus k vertical line X greater than x close parentheses equals P open parentheses X equals k close parentheses where x and k are positive integers

    • If the first x trials fail, then ignore them and treat trial open parentheses x plus 1 close parentheses as the first trial

Examiner Tips and Tricks

Some graphical calculators have functions to calculate probabilities using a geometric distribution. The functions work the same way as those for a binomial distribution. However, it is just as easy to calculate probabilities for a geometric distribution using the formula as a calculator.

Worked Example

On a gaming app, a player gets a free spin of a roulette wheel every day. There is a 2% chance that the spin of the roulette wheel results in the player winning a prize. The result of each spin is independent of any previous results.

(a) Find the probability that the player first wins a prize on the 8th day.

Answer:

The number of days until the first win can be modeled by a geometric distribution

Define the variable and state the distribution

Let X be the number of the day that the player first wins a prize by spinning the roulette wheel

X is a geometric random variable with p equals 0.02

The required probability is P open parentheses X equals 8 close parentheses

Use the formula P open parentheses X equals x close parentheses equals open parentheses 1 minus p close parentheses to the power of x minus 1 end exponent p to calculate the probability

table row cell P open parentheses X equals 8 close parentheses end cell equals cell open parentheses 1 minus 0.02 close parentheses to the power of 8 minus 1 end exponent open parentheses 0.02 close parentheses end cell row blank equals cell open parentheses 0.98 close parentheses to the power of 7 open parentheses 0.02 close parentheses end cell row blank equals cell 0.01736... end cell end table

The probability that the player first wins a prize on the 8th day is 0.0174

(b) Find the probability that the player does not win a prize in the first 10 days.

Answer:

The distribution is the same as in the previous part

Therefore, the variable does not need to be stated again

The required probability is P open parentheses X greater than 10 close parentheses which is the same as P open parentheses X greater or equal than 11 close parentheses for a geometric distribution

Use the formula P open parentheses X greater or equal than x close parentheses equals open parentheses 1 minus p close parentheses to the power of x minus 1 end exponent to calculate the probability

Note that this is the same as finding the probability of not winning a prize on each of the ten days

table row cell P open parentheses X greater than 10 close parentheses end cell equals cell open parentheses 1 minus 0.02 close parentheses to the power of 11 minus 1 end exponent end cell row blank equals cell 0.98 to the power of 10 end cell row blank equals cell 0.8170... end cell end table

The probability that the player does not win a prize in the first ten days is 0.817

(c) Given that the player has not won a prize in the first ten days, find the probability that they first win a prize within the first 15 days.

Answer:

The required probability is P open parentheses X less or equal than 15 vertical line X greater than 10 close parentheses

The memoryless property can be applied

The first 10 days can be ignored, as they are failures, and counting can start from day 11

This means the probability is equal to the probability that the player first wins a prize within the first 5 days

Use the formula P open parentheses X less or equal than x close parentheses equals 1 minus open parentheses 1 minus p close parentheses to the power of x

Note that this is the same as finding the probability that the player does not fail to win a prize on each of the five days

table row cell P open parentheses X less or equal than 15 vertical line X greater than 10 close parentheses end cell equals cell P open parentheses X less or equal than 5 close parentheses end cell row blank equals cell 1 minus open parentheses 1 minus 0.02 close parentheses to the power of 5 end cell row blank equals cell 1 minus 0.98 to the power of 5 end cell row blank equals cell 0.09607... end cell end table

The probability that the player first wins a prize within the first 15 days given that they have not won a prize in the first 10 days is 0.0961

Last updated:

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Introduction to Geometric DistributionsNext:Introduction to Sampling Distributions
    Author
    Reviewer
    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

    Lucy Kirkham

    Author: Lucy Kirkham

    Expertise: Head of STEM

    Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.