Conditional Probability

What you'll learn

How to read event notation such as $A^{'}$ , $\cap B$ and $\cup B$ .
How to calculate probabilities when you are told something has already happened.
How to test whether two events are independent.
How to use tree diagrams and Venn diagrams for “given that” questions.

1. Events, complements, intersections and unions

In probability, letters such as $A$ and $B$ usually represent events. Probabilities are always between 0 and 1, and the total probability of all possible outcomes is 1.

Definition

Core event notation

An event is a set of outcomes, such as “a selected person watches football”.
The complement of $A$ , written $A^{'}$ , means $A$ does not happen.
The intersection $\cap B$ means both $A$ and $B$ happen.
The union $\cup B$ means at least one of $A$ or $B$ happens, including both.

For two events, the key “or” formula is:

\cup B)=P(A)+P(B)-P(A \cap B)

You subtract the overlap because it has been counted twice.

A two-event Venn diagram showing why the overlap must be subtracted once when finding a union.

Example

Finding a union and the outside region

Events $A$ and $B$ have $P (A) = 0.45$ , $P (B) = 0.50$ and $\cap B)=0.18$ . Find $\cup B)$ and $\cap B')$ .

Use the union formula.
$\cup B)=0.45+0.50-0.18=0.77$
The event $\cap B'$ means neither $A$ nor $B$ , so it is outside the union.
$\cap B')=1-P(A \cup B)=1-0.77=0.23$

Common Mistake

Forgetting the overlap

If you calculate $P (A) + P (B)$ for “ $A$ or $B$ ”, you count outcomes in both events twice. Always subtract $\cap B)$ once.

2. Conditional probability: “given that”

Conditional probability is about updating the probability when you know extra information.

Definition

Conditional probability

$P(A∣B)P(A\mid B)$ means “the probability of $A$ , given that $B$ has happened”. Provided $P (B) > 0$ ,

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A \cap B)}{P(B)}

Key Idea

Shrink the universe

When you see “given $B$ ”, pretend $B$ is the whole universe. The denominator is $P (B)$ .

Example

Using the conditional probability formula

Events $A$ and $B$ have $P (A) = 0.48$ , $P (B) = 0.60$ and $\cap B)=0.30$ . Find $P(A∣B)P(A\mid B)$ and $P(A′∣B′)P(A'\mid B')$ .

A Venn diagram partitioning the sample space into the regions needed for the conditional probabilities in the example.

For $P(A∣B)P(A\mid B)$ , divide the overlap by the probability of the condition.
$P(A∣B)=0.300.60=0.5P(A\mid B)=\frac{0.30}{0.60}=0.5$
For $P(A′∣B′)P(A'\mid B')$ , first find the union.
$\cup B)=0.48+0.60-0.30=0.78$
The outside-both region is $\cap B'$ .
$\cap B')=1-0.78=0.22$
Now condition on $B^{'}$ , where $P (B^{'}) = 1 - 0.60 = 0.40$ .
$P(A′∣B′)=P(A′∩B′)P(B′)=0.220.40=0.55P(A'\mid B')=\frac{P(A' \cap B')}{P(B')}=\frac{0.22}{0.40}=0.55$

Tip

Read the bar out loud

In $P(A∣B)P(A\mid B)$ , say “ $A$ given $B$ ”. The event after the bar is what you divide by.

3. Rearranging the conditional probability formula

Sometimes you are given a conditional probability and need to work backwards to find an intersection.

From

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A \cap B)}{P(B)}

we get:

\cap B)=P(A\mid B)P(B)

Example

Working backwards from a conditional probability

Events $A$ and $B$ have $P(A)=518P(A)=\frac{5}{18}$ , $P(B)=23P(B)=\frac{2}{3}$ and $P(A∣B)=14P(A\mid B)=\frac{1}{4}$ . Find $\cap B)$ , $P(B′∣A)P(B'\mid A)$ and $\cup B)$ .

$A Venn diagram showing the fraction in each region after working backwards from the conditional probability.$

Multiply to find the intersection.
$\cap B)=P(A\mid B)P(B)=\frac{1}{4}\times \frac{2}{3}=\frac{1}{6}$
To find $P(B′∣A)P(B'\mid A)$ , first find the part of $A$ that is not in $B$ .
$\cap B')=P(A)-P(A \cap B)=\frac{5}{18}-\frac{1}{6}=\frac{1}{9}$
Now divide by the condition, $P (A)$ .
$P(B′∣A)=P(A∩B′)P(A)=19518=25P(B'\mid A)=\frac{P(A \cap B')}{P(A)}=\frac{\frac{1}{9}}{\frac{5}{18}}=\frac{2}{5}$
For $\cup B)$ , use the complement. The only excluded region is $\cap B'$ .
$\cup B)=1-P(A \cap B')=1-\frac{1}{9}=\frac{8}{9}$

Common Mistake

Using the wrong denominator

$P(B′∣A)P(B'\mid A)$ is not divided by $P (B^{'})$ . The phrase “given $A$ ” means divide by $P (A)$ .

4. Independence and mutually exclusive events

Two important phrases sound similar but mean very different things.

A side-by-side comparison of independent overlapping events and mutually exclusive events.

Definition

Independent and mutually exclusive

Events $A$ and $B$ are independent if knowing one happened does not change the probability of the other. For independent events, $\cap B)=P(A)P(B)$ .
Events $A$ and $B$ are mutually exclusive if they cannot happen together. For mutually exclusive events, $\cap B)=0$ .

Common Mistake

Do not mix these up

If two events with non-zero probabilities are mutually exclusive, they are not independent: knowing one happened makes the other impossible.

Example

Using independence to find a missing probability

Suppose $P (A) = 0.30$ , $\cup B)=0.72$ , and $A$ and $B$ are independent. Find $P (B)$ and $\cup B')$ .

Let $P (B) = x$ . Since the events are independent, $\cap B)=0.30x$ .
Substitute into the union formula.
$0.72 = 0.30 + x - 0.30 x$
Solve for $x$ .
$\Rightarrow x=0.60$
$\cup B'$ means “not both $A$ and $B$ ”, so use the complement of $\cap B$ .
$\cup B')=1-P(A \cap B)=1-(0.30\times 0.60)=0.82$

5. Tree diagrams and reverse conditional probability

Tree diagrams are especially useful when there are stages, such as choosing a machine then checking whether an item is faulty, or removing counters one after another.

Definition

Tree diagram

A tree diagram shows probabilities on branches in stages. Multiply along a complete path; add probabilities of separate paths that lead to the same final outcome.

When you are told the final outcome and asked about an earlier branch, use the same conditional probability idea:

event)P(\text{earlier event}\mid \text{final event})=\frac{P(\text{earlier event and final event})}{P(\text{final event})}

Example

Faulty components from different machines

A factory uses three machines. Machine $A$ makes 50% of the components, machine $B$ makes 30%, and machine $C$ makes 20%. Their faulty rates are 4%, 2% and 5% respectively. Find the probability a component is faulty, and then find the probability it came from machine $C$ given that it is faulty.

A probability tree for the machine and faulty-component example, with the faulty paths highlighted for the reverse conditional probability.

Multiply along each “faulty” path.
$P(A∩F)=0.50×0.04=0.020P(B∩F)=0.30×0.02=0.006P(C∩F)=0.20×0.05=0.010\begin{aligned} P(A \cap F)&=0.50\times 0.04=0.020\\ P(B \cap F)&=0.30\times 0.02=0.006\\ P(C \cap F)&=0.20\times 0.05=0.010 \end{aligned}$
Add the separate faulty paths.
$P (F) = 0.020 + 0.006 + 0.010 = 0.036$
Given faulty, restrict to the faulty outcomes.
$P(C∣F)=P(C∩F)P(F)=0.0100.036=518P(C\mid F)=\frac{P(C \cap F)}{P(F)}=\frac{0.010}{0.036}=\frac{5}{18}$

Tip

Without replacement

If an item is not put back, update both the favourable count and the total count on the next branch. After taking one counter from 12 counters, there are 11 counters left.

6. Venn diagrams with three events

A Venn diagram is a picture showing how events overlap. With three events, fill the most specific regions first: usually the “all three” overlap, then the pairwise-only overlaps, then the single-event-only regions.

Example

A three-event Venn diagram with sports

A group of 60 students are asked whether they watch football, golf and hockey. Six watch all three. The totals for football and golf, golf and hockey, and football and hockey are 14, 10 and 12 respectively. Overall, 35 watch football. Given a student watches football, find the probability they do not watch hockey.

A three-event Venn diagram filled from the centre outwards for the football, golf and hockey example.

Pair totals include the all-three region, so subtract 6 to get the pairwise-only regions.
$only=12−6=6\begin{aligned} \text{football and golf only} &= 14-6=8\\ \text{golf and hockey only} &= 10-6=4\\ \text{football and hockey only} &= 12-6=6 \end{aligned}$
Find the football-only region.
$only=35−(8+6+6)=15\text{football only}=35-(8+6+6)=15$
“Football but not hockey” includes football only and football-and-golf only.
$hockey∣F)=15+835=2335P(\text{not hockey}\mid F)=\frac{15+8}{35}=\frac{23}{35}$

Tip

Pair totals include the middle

If a question says 14 watch football and golf, that usually includes students who also watch hockey. Subtract the all-three region to get the “football and golf only” region.

Exam technique

In the exam

Underline the words “given that” and write the denominator first.
For “or”, use $\cup B)=P(A)+P(B)-P(A \cap B)$ or add disjoint Venn regions.
For tree diagrams, multiply along branches and add separate paths.
For independence, check whether $\cap B)=P(A)P(B)$ or whether $P(A∣B)=P(A)P(A\mid B)=P(A)$ .
For “at least one”, consider using the complement “none”.

Self review

Check yourself

In $P(A′∣B)P(A'\mid B)$ , which probability goes in the denominator?
How can you find $\cap B')$ if you know $\cup B)$ ?
What calculation would show that two events are independent?

Conditional Probability

What you'll learn

How to read event notation such as $A^{'}$ , $\cap B$ and $\cup B$ .
How to calculate probabilities when you are told something has already happened.
How to test whether two events are independent.
How to use tree diagrams and Venn diagrams for “given that” questions.

1. Events, complements, intersections and unions

In probability, letters such as $A$ and $B$ usually represent events. Probabilities are always between 0 and 1, and the total probability of all possible outcomes is 1.

Definition

Core event notation

An event is a set of outcomes, such as “a selected person watches football”.
The complement of $A$ , written $A^{'}$ , means $A$ does not happen.
The intersection $\cap B$ means both $A$ and $B$ happen.
The union $\cup B$ means at least one of $A$ or $B$ happens, including both.

For two events, the key “or” formula is:

\cup B)=P(A)+P(B)-P(A \cap B)

You subtract the overlap because it has been counted twice.

A two-event Venn diagram showing why the overlap must be subtracted once when finding a union.

Example

Finding a union and the outside region

Events $A$ and $B$ have $P (A) = 0.45$ , $P (B) = 0.50$ and $\cap B)=0.18$ . Find $\cup B)$ and $\cap B')$ .

Use the union formula.
$\cup B)=0.45+0.50-0.18=0.77$
The event $\cap B'$ means neither $A$ nor $B$ , so it is outside the union.
$\cap B')=1-P(A \cup B)=1-0.77=0.23$

Common Mistake

Forgetting the overlap

If you calculate $P (A) + P (B)$ for “ $A$ or $B$ ”, you count outcomes in both events twice. Always subtract $\cap B)$ once.

2. Conditional probability: “given that”

Conditional probability is about updating the probability when you know extra information.

Definition

Conditional probability

$P(A∣B)P(A\mid B)$ means “the probability of $A$ , given that $B$ has happened”. Provided $P (B) > 0$ ,

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A \cap B)}{P(B)}

Key Idea

Shrink the universe

When you see “given $B$ ”, pretend $B$ is the whole universe. The denominator is $P (B)$ .

Example

Using the conditional probability formula

Events $A$ and $B$ have $P (A) = 0.48$ , $P (B) = 0.60$ and $\cap B)=0.30$ . Find $P(A∣B)P(A\mid B)$ and $P(A′∣B′)P(A'\mid B')$ .

A Venn diagram partitioning the sample space into the regions needed for the conditional probabilities in the example.

For $P(A∣B)P(A\mid B)$ , divide the overlap by the probability of the condition.
$P(A∣B)=0.300.60=0.5P(A\mid B)=\frac{0.30}{0.60}=0.5$
For $P(A′∣B′)P(A'\mid B')$ , first find the union.
$\cup B)=0.48+0.60-0.30=0.78$
The outside-both region is $\cap B'$ .
$\cap B')=1-0.78=0.22$
Now condition on $B^{'}$ , where $P (B^{'}) = 1 - 0.60 = 0.40$ .
$P(A′∣B′)=P(A′∩B′)P(B′)=0.220.40=0.55P(A'\mid B')=\frac{P(A' \cap B')}{P(B')}=\frac{0.22}{0.40}=0.55$

Tip

Read the bar out loud

In $P(A∣B)P(A\mid B)$ , say “ $A$ given $B$ ”. The event after the bar is what you divide by.

3. Rearranging the conditional probability formula

Sometimes you are given a conditional probability and need to work backwards to find an intersection.

From

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A \cap B)}{P(B)}

we get:

\cap B)=P(A\mid B)P(B)

Example

Working backwards from a conditional probability

Events $A$ and $B$ have $P(A)=518P(A)=\frac{5}{18}$ , $P(B)=23P(B)=\frac{2}{3}$ and $P(A∣B)=14P(A\mid B)=\frac{1}{4}$ . Find $\cap B)$ , $P(B′∣A)P(B'\mid A)$ and $\cup B)$ .

$A Venn diagram showing the fraction in each region after working backwards from the conditional probability.$

Multiply to find the intersection.
$\cap B)=P(A\mid B)P(B)=\frac{1}{4}\times \frac{2}{3}=\frac{1}{6}$
To find $P(B′∣A)P(B'\mid A)$ , first find the part of $A$ that is not in $B$ .
$\cap B')=P(A)-P(A \cap B)=\frac{5}{18}-\frac{1}{6}=\frac{1}{9}$
Now divide by the condition, $P (A)$ .
$P(B′∣A)=P(A∩B′)P(A)=19518=25P(B'\mid A)=\frac{P(A \cap B')}{P(A)}=\frac{\frac{1}{9}}{\frac{5}{18}}=\frac{2}{5}$
For $\cup B)$ , use the complement. The only excluded region is $\cap B'$ .
$\cup B)=1-P(A \cap B')=1-\frac{1}{9}=\frac{8}{9}$

Common Mistake

Using the wrong denominator

$P(B′∣A)P(B'\mid A)$ is not divided by $P (B^{'})$ . The phrase “given $A$ ” means divide by $P (A)$ .

4. Independence and mutually exclusive events

Two important phrases sound similar but mean very different things.

A side-by-side comparison of independent overlapping events and mutually exclusive events.

Definition

Independent and mutually exclusive

Events $A$ and $B$ are independent if knowing one happened does not change the probability of the other. For independent events, $\cap B)=P(A)P(B)$ .
Events $A$ and $B$ are mutually exclusive if they cannot happen together. For mutually exclusive events, $\cap B)=0$ .

Common Mistake

Do not mix these up

If two events with non-zero probabilities are mutually exclusive, they are not independent: knowing one happened makes the other impossible.

Example

Using independence to find a missing probability

Suppose $P (A) = 0.30$ , $\cup B)=0.72$ , and $A$ and $B$ are independent. Find $P (B)$ and $\cup B')$ .

Let $P (B) = x$ . Since the events are independent, $\cap B)=0.30x$ .
Substitute into the union formula.
$0.72 = 0.30 + x - 0.30 x$
Solve for $x$ .
$\Rightarrow x=0.60$
$\cup B'$ means “not both $A$ and $B$ ”, so use the complement of $\cap B$ .
$\cup B')=1-P(A \cap B)=1-(0.30\times 0.60)=0.82$

5. Tree diagrams and reverse conditional probability

Tree diagrams are especially useful when there are stages, such as choosing a machine then checking whether an item is faulty, or removing counters one after another.

Definition

Tree diagram

A tree diagram shows probabilities on branches in stages. Multiply along a complete path; add probabilities of separate paths that lead to the same final outcome.

When you are told the final outcome and asked about an earlier branch, use the same conditional probability idea:

event)P(\text{earlier event}\mid \text{final event})=\frac{P(\text{earlier event and final event})}{P(\text{final event})}

Example

Faulty components from different machines

A probability tree for the machine and faulty-component example, with the faulty paths highlighted for the reverse conditional probability.

Multiply along each “faulty” path.
$P(A∩F)=0.50×0.04=0.020P(B∩F)=0.30×0.02=0.006P(C∩F)=0.20×0.05=0.010\begin{aligned} P(A \cap F)&=0.50\times 0.04=0.020\\ P(B \cap F)&=0.30\times 0.02=0.006\\ P(C \cap F)&=0.20\times 0.05=0.010 \end{aligned}$
Add the separate faulty paths.
$P (F) = 0.020 + 0.006 + 0.010 = 0.036$
Given faulty, restrict to the faulty outcomes.
$P(C∣F)=P(C∩F)P(F)=0.0100.036=518P(C\mid F)=\frac{P(C \cap F)}{P(F)}=\frac{0.010}{0.036}=\frac{5}{18}$

Tip

Without replacement

If an item is not put back, update both the favourable count and the total count on the next branch. After taking one counter from 12 counters, there are 11 counters left.

6. Venn diagrams with three events

Example

A three-event Venn diagram with sports

A three-event Venn diagram filled from the centre outwards for the football, golf and hockey example.

Pair totals include the all-three region, so subtract 6 to get the pairwise-only regions.
$only=12−6=6\begin{aligned} \text{football and golf only} &= 14-6=8\\ \text{golf and hockey only} &= 10-6=4\\ \text{football and hockey only} &= 12-6=6 \end{aligned}$
Find the football-only region.
$only=35−(8+6+6)=15\text{football only}=35-(8+6+6)=15$
“Football but not hockey” includes football only and football-and-golf only.
$hockey∣F)=15+835=2335P(\text{not hockey}\mid F)=\frac{15+8}{35}=\frac{23}{35}$

Tip

Pair totals include the middle

If a question says 14 watch football and golf, that usually includes students who also watch hockey. Subtract the all-three region to get the “football and golf only” region.

Exam technique

In the exam

Underline the words “given that” and write the denominator first.
For “or”, use $\cup B)=P(A)+P(B)-P(A \cap B)$ or add disjoint Venn regions.
For tree diagrams, multiply along branches and add separate paths.
For independence, check whether $\cap B)=P(A)P(B)$ or whether $P(A∣B)=P(A)P(A\mid B)=P(A)$ .
For “at least one”, consider using the complement “none”.

Self review

Check yourself

In $P(A′∣B)P(A'\mid B)$ , which probability goes in the denominator?
How can you find $\cap B')$ if you know $\cup B)$ ?
What calculation would show that two events are independent?

Conditional Probability

What you'll learn

1. Events, complements, intersections and unions

2. Conditional probability: “given that”

3. Rearranging the conditional probability formula

4. Independence and mutually exclusive events

5. Tree diagrams and reverse conditional probability

6. Venn diagrams with three events

Recap questions

You've reached the end

Conditional Probability

What you'll learn

1. Events, complements, intersections and unions

2. Conditional probability: “given that”

3. Rearranging the conditional probability formula

4. Independence and mutually exclusive events

5. Tree diagrams and reverse conditional probability

6. Venn diagrams with three events

Recap questions

You've reached the end

Conditional Probability

What you'll learn

1. Events, complements, intersections and unions

2. Conditional probability: “given that”

3. Rearranging the conditional probability formula

4. Independence and mutually exclusive events

5. Tree diagrams and reverse conditional probability

6. Venn diagrams with three events

Recap questions

You've reached the end

Conditional Probability Revision Guide

Conditional Probability

What you'll learn

1. Events, complements, intersections and unions

2. Conditional probability: “given that”

3. Rearranging the conditional probability formula

4. Independence and mutually exclusive events

5. Tree diagrams and reverse conditional probability

6. Venn diagrams with three events

Recap questions

You've reached the end

Conditional Probability Revision Guide