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AP®StatisticsCollege BoardStudy GuidesUnit 4: Probability, Random Variables & Probability DistributionsProbabilityProbabilities of Combined Events using the Rules

Probabilities of Combined Events using the Rules (College Board AP® Statistics)

Study Guide

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Probabilities of combined events

What are the useful rules for calculating probabilities of combined events?

  • You should be able to use the following rules:

Rule

Formula

Complementary events formula

P open parentheses A apostrophe close parentheses equals 1 minus P open parentheses A close parentheses

Addition rule

P open parentheses A union B close parentheses equals P open parentheses A close parentheses plus P open parentheses B close parentheses minus P open parentheses A intersection B close parentheses

Multiplication rule

P open parentheses A intersection B close parentheses equals P open parentheses A close parentheses times P open parentheses B vertical line A close parentheses

Partition rule

P open parentheses B close parentheses equals P open parentheses A intersection B close parentheses plus P open parentheses A apostrophe intersection B close parentheses

Conditional probability formula

P open parentheses A vertical line B close parentheses equals fraction numerator P open parentheses A intersection B close parentheses over denominator P open parentheses B close parentheses end fraction

  • The conditional probability formula can be rewritten so that it can be used in various scenarios depending on which probabilities you are given in a question

    • Using the partition rule on the denominator gives you

      • P open parentheses A vertical line B close parentheses equals fraction numerator P open parentheses A intersection B close parentheses over denominator P open parentheses A intersection B close parentheses plus P open parentheses A apostrophe intersection B close parentheses end fraction

    • Using the multiplication rule on each term gives you

      • P open parentheses A vertical line B close parentheses equals fraction numerator P open parentheses A close parentheses times P open parentheses B vertical line A close parentheses over denominator P open parentheses A close parentheses times P open parentheses B vertical line A close parentheses plus P open parentheses A apostrophe close parentheses times P open parentheses B vertical line A apostrophe close parentheses end fraction

      • This is called Bayes' theorem

Examiner Tips and Tricks

Exam questions tend to help you through trickier problems. For example, they might ask you to find P open parentheses B close parentheses before finding P open parentheses A vertical line B close parentheses.

How can I calculate probabilities without using a tree diagram?

  • Rewrite the event using the possible outcomes and the words "not", "and" and "or"

    • The word "not" means you subtract the probability from 1

    • The word "or" means you add the probabilities

      • This is because the outcomes are mutually exclusive

    • The word "and" means you multiply the probabilities

      • This might involve conditional probabilities

    • For example, consider the following events when three counters are chosen from a bag

      • B subscript 1, B subscript 2 and B subscript 3 are the events that the first, second and third counter chosen is blue, respectively

Event

Rephrased

Calculation

All the chosen counters are blue

Blue and blue and blue

P open parentheses B subscript 1 intersection B subscript 2 intersection B subscript 3 close parentheses

Exactly two chosen counters are blue

Blue and blue and not blue
or

blue and not blue and blue

or

not blue and blue and blue

P open parentheses B subscript 1 intersection B subscript 2 intersection B subscript 3 apostrophe close parentheses plus P open parentheses B subscript 1 intersection B subscript 2 apostrophe intersection B subscript 3 close parentheses plus P open parentheses B subscript 1 apostrophe intersection B subscript 2 intersection B subscript 3 close parentheses

At least one chosen counter is blue

Not:

not blue and not blue and not blue

1 minus P open parentheses B subscript 1 apostrophe intersection B subscript 2 apostrophe intersection B subscript 3 apostrophe close parentheses

How do I find probabilities of independent events?

  • To find the probability of two independent events occurring at the same time

    • multiply the probability of one event occurring by the probability of the other event occurring

  • This can be extended to any number of independent events by multiplying the probabilities of all the events together

    • This is useful when a trial is repeated multiple times

      • e.g. the probability that a fair coin always lands on tails when flipped 5 times is 1 half times 1 half times 1 half times 1 half times 1 half equals open parentheses 1 half close parentheses to the power of 5

  • Check to see if there are multiple outcomes for an event

    • e.g. there are six ways of selecting a blue, red and yellow counter when selecting three counters from a bag

      • BRY, BYR, RBY, RYB, RBY, RYB

      • If the probabilities of each outcome are equal then you can just multiply the probability of one outcome by six

Worked Example

Three friends, Rachel, Monica and Phoebe, live in separate apartments. The probabilities of each person getting a visitor to their apartment on any given day are 0.7, 0.6 and 0.1 respectively. A friend getting a visitor is independent of the other friends getting visitors.

What is the probability that at least one of the three friends gets a visitor to their apartment on a random day?

Answer:

Let R represent the event that Rachel gets a visitor

Let M represent the event that Monica gets a visitor

Let B represent the event that Phoebe gets a visitor

At least one of the friends getting a visitor has multiple options including R intersection M intersection B , R intersection M intersection B apostrophe , R intersection M apostrophe intersection B, etc.

It is quicker to consider the event where none of the friends get a visitor as this is the complementary event

A friend getting a visitor is independent of either of the other friends getting a visitor, therefore you can find the probability by multiplying the probabilities of not getting a visitor together

table row cell P open parentheses R apostrophe intersection M apostrophe intersection B apostrophe close parentheses end cell equals cell open parentheses 1 minus 0.7 close parentheses times open parentheses 1 minus 0.6 close parentheses times open parentheses 1 minus 0.1 close parentheses end cell row blank equals cell 0.3 times 0.4 times 0.9 end cell row blank equals cell 0.108 end cell end table

Subtract this from 1

1 minus 0.108 equals 0.892

The probability that at least one of the three friends gets a visitor to their apartment on a random day is 0.892

Worked Example

Joey works at a coffee house where customers can either order a cappuccino or a latte. 70% of customers order cappuccinos and the rest order lattes. There is a 60% chance that Joey gets the order incorrect if a customer orders a cappuccino. There is a 25% chance that Joey gets the order incorrect if a customer orders a latte.

(a) What is the probability that Joey gets a customer's order incorrect?

Answer:

Let I represent the event that Joey gets the order incorrect

Let C represent the event that the customer orders a cappuccino

Let L represent the event that the customer orders a latte

Identify the probabilities given in the question

P open parentheses C close parentheses equals 0.7 P open parentheses L close parentheses equals 0.3 P open parentheses I vertical line C close parentheses equals 0.6 P open parentheses I vertical line L close parentheses equals 0.25

There are two outcomes that involve Joey getting the order incorrect

The customer ordered a cappuccino or the customer ordered a latte

Use the partition rule to write the probability that Joey gets the order incorrect as the sum of the probabilities of the two relevant outcomes

P open parentheses I close parentheses equals P open parentheses C intersection I close parentheses plus P open parentheses L intersection I close parentheses

The conditional probabilities for Joey getting the order incorrect are known for each drink

Rewrite the probability of each outcome using the multiplication rule

P open parentheses I close parentheses equals P open parentheses C close parentheses times P open parentheses I vertical line C close parentheses plus P open parentheses L close parentheses times P open parentheses I vertical line L close parentheses

Substitute the probabilities into the formula

table row cell P open parentheses I close parentheses end cell equals cell 0.7 times 0.6 plus 0.3 times 0.25 end cell row blank equals cell 0.495 end cell end table

The probability that Joey gets a customer's order incorrect is 0.495

(b) Given that Joey gets a customer's order incorrect, what is the probability that they ordered a cappuccino?

Answer:

The required probability is P open parentheses C vertical line I close parentheses

To use the conditional probability formula P open parentheses C intersection I close parentheses is needed

Rewrite this probability using the multiplication rule in the same way as the previous part

table row cell P open parentheses C intersection I close parentheses end cell equals cell P open parentheses C close parentheses times P open parentheses I vertical line C close parentheses end cell row blank equals cell 0.7 times 0.6 end cell row blank equals cell 0.42 end cell end table

Use the conditional probability formula P open parentheses C vertical line I close parentheses equals fraction numerator P open parentheses C intersection I close parentheses over denominator P open parentheses I close parentheses end fraction and the answer from the previous part

table row cell P open parentheses C vertical line I close parentheses end cell equals cell fraction numerator 0.42 over denominator 0.495 end fraction end cell row blank equals cell 28 over 33 end cell row blank equals cell 0.8484... end cell end table

Given that Joey gets a customer's order incorrect, the probability they ordered a cappuccino is 28 over 33 (roughly 0.848)

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    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.