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

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Dan Finlay

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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	$P (A') = 1 - P (A)$
Addition rule	$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Multiplication rule	$P (A \cap B) = P (A) \cdot P (B \| A)$
Partition rule	$P (B) = P (A \cap B) + P (A' \cap B)$
Conditional probability formula	$P (A \| B) = \frac{P (A \cap B)}{P (B)}$

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
  - $P (A | B) = \frac{P (A \cap B)}{P (A \cap B) + P (A' \cap B)}$
- Using the multiplication rule on each term gives you
  - $P (A | B) = \frac{P (A) \cdot P (B | A)}{P (A) \cdot P (B | A) + P (A') \cdot P (B | A')}$
  - This is called Bayes' theorem

Exam Tip

Exam questions tend to help you through trickier problems. For example, they might ask you to find $P (B)$ before finding $P (A | B)$ .

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
  - $B_{1}$ , $B_{2}$ and $B_{3}$ 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	$P (B_{1} \cap B_{2} \cap B_{3})$
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	$P (B_{1} \cap B_{2} \cap B_{3}') + P (B_{1} \cap B_{2}' \cap B_{3}) + P (B_{1}' \cap B_{2} \cap B_{3})$
At least one chosen counter is blue	Not: not blue and not blue and not blue	$1 - P (B_{1}' \cap B_{2}' \cap B_{3}')$

Event

Rephrased

Calculation

All the chosen counters are blue

Blue and blue and blue

$P (B_{1} \cap B_{2} \cap B_{3})$

Exactly two chosen counters are blue

Blue and blue and not blue
or

blue and not blue and blue

not blue and blue and blue

$P (B_{1} \cap B_{2} \cap B_{3}') + P (B_{1} \cap B_{2}' \cap B_{3}) + P (B_{1}' \cap B_{2} \cap B_{3})$

At least one chosen counter is blue

Not:

not blue and not blue and not blue

$1 - P (B_{1}' \cap B_{2}' \cap B_{3}')$

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 $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = {(\frac{1}{2})}^{5}$
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 $R \cap M \cap B$ , $R \cap M \cap B'$ , $R \cap M' \cap B$ , 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

$\begin{array}{rcl} P (R' \cap M' \cap B') & = & (1 - 0.7) \cdot (1 - 0.6) \cdot (1 - 0.1) \\ = & 0.3 \cdot 0.4 \cdot 0.9 \\ = & 0.108 \end{array}$

Subtract this from 1

$1 - 0.108 = 0.892$

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

$P (C) = 0.7 P (L) = 0.3 P (I | C) = 0.6 P (I | L) = 0.25$

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

$P (I) = P (C \cap I) + P (L \cap I)$

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

Rewrite the probability of each outcome using the multiplication rule

$P (I) = P (C) \cdot P (I | C) + P (L) \cdot P (I | L)$

Substitute the probabilities into the formula

$\begin{array}{rcl} P (I) & = & 0.7 \cdot 0.6 + 0.3 \cdot 0.25 \\ = & 0.495 \end{array}$

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 $P (C | I)$

To use the conditional probability formula $P (C \cap I)$ is needed

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

$\begin{array}{rcl} P (C \cap I) & = & P (C) \cdot P (I | C) \\ = & 0.7 \cdot 0.6 \\ = & 0.42 \end{array}$

Use the conditional probability formula $P (C | I) = \frac{P (C \cap I)}{P (I)}$ and the answer from the previous part

$\begin{array}{rcl} P (C | I) & = & \frac{0.42}{0.495} \\ = & \frac{28}{33} \\ = & 0.8484 . . . \end{array}$

Given that Joey gets a customer's order incorrect, the probability they ordered a cappuccino is $\frac{28}{33}$ (roughly 0.848)

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

Revision Note

Probabilities of combined events

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

Exam Tip

How can I calculate probabilities without using a tree diagram?

How do I find probabilities of independent events?

Worked Example

Worked Example

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Author: Dan Finlay