2.3.4 Conditional probability from first principles (A-level only)

What you'll learn

What it means for a probability to be conditional on an event.
How to restrict a sample space when new information is given.
How to calculate conditional probabilities directly from outcomes, frequencies and set information.
How to distinguish conditional probability from intersection and independence.

Prerequisites: events and intersections

A sample space is the set of all possible outcomes of an experiment. An event is a set of outcomes with a particular property.

For example, when rolling a fair six-sided die, the sample space is

S=\{1,2,3,4,5,6\}.

If event $A$ is “the result is even”, then

A=\{2,4,6\}.

If $A$ and $B$ are two events, the intersection $A∩BA\cap B$ contains the outcomes that belong to both events.

For instance, if $B$ is “the result is greater than 3”, then

B=\{4,5,6\}

and

A∩B={4,6}.A\cap B=\{4,6\}.

Definition

Intersection of events

The event $A∩BA\cap B$ occurs when both $A$ and $B$ occur.

What is conditional probability?

Sometimes you are given extra information about the outcome of an experiment. This information changes which outcomes are still possible.

The conditional probability of $A$ given $B$ is the probability that $A$ occurs when you already know that $B$ has occurred. It is written as $P(A∣B)P(A\mid B)$ .

The vertical line means “given that”, so you read $P(A∣B)P(A\mid B)$ as:

the probability of $A$ , given that $B$ has occurred.

Definition

Conditional probability

Provided $P (B) > 0$ , the conditional probability of $A$ given $B$ is

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

The event after the vertical line is the condition. It becomes your new, restricted sample space.

A Venn diagram showing the original sample space and how conditioning on B restricts attention to B, with the intersection of A and B highlighted

Key Idea

Restrict, then compare

To find $P(A∣B)P(A\mid B)$ , restrict attention to outcomes in $B$ , then find the proportion of those outcomes that also belong to $A$ .

Restricting a die sample space

Example

Conditioning on an even die score

A fair die is rolled. Let $A$ be the event “the score is greater than 3” and let $B$ be the event “the score is even”. Find $P(A∣B)P(A\mid B)$ .

Knowing that $B$ occurred restricts the possible outcomes to
$B=\{2,4,6\}.$
Within this restricted sample space, the outcomes greater than 3 are 4 and 6. Therefore,
$A∩B={4,6}.A\cap B=\{4,6\}.$
Two of the three equally likely outcomes in $B$ also belong to $A$ , so
$P(A∣B)=23.P(A\mid B)=\frac{2}{3}.$

Notice that the original sample space had six outcomes, but the denominator is now 3 because the condition tells you the result is even.

Common Mistake

Using the original denominator

Once you are told that $B$ has occurred, outcomes outside $B$ are no longer possible. The denominator must represent the outcomes in $B$ , not the whole original sample space.

Deriving the formula from first principles

Suppose a sample space contains $n (S)$ equally likely outcomes, where $n (X)$ means the number of outcomes in set $X$ .

After conditioning on $B$ , there are $n (B)$ outcomes under consideration. Of these, $n(A∩B)n(A\cap B)$ also satisfy $A$ . Therefore,

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

For equally likely outcomes,

P(A∩B)=n(A∩B)n(S)P(A\cap B)=\frac{n(A\cap B)}{n(S)}

and

P(B)=n(B)n(S).P(B)=\frac{n(B)}{n(S)}.

Dividing these probabilities gives

P(A∩B)P(B)=n(A∩B)n(S)n(B)n(S)=n(A∩B)n(B).\begin{aligned} \frac{P(A\cap B)}{P(B)} &=\frac{\frac{n(A\cap B)}{n(S)}}{\frac{n(B)}{n(S)}}\\ &=\frac{n(A\cap B)}{n(B)}. \end{aligned}

This produces the conditional probability formula directly.

Common Mistake

The condition must be possible

The expression $P(A∣B)P(A\mid B)$ is not defined when $P (B) = 0$ , because the formula would require division by zero.

Calculating from a frequency table

In real contexts, you may be given observed frequencies rather than theoretical probabilities. The same principle applies:

Select the row, column or group named after “given that”.
Use the total frequency of that group as the denominator.
Count how many members of that group also satisfy the required event.

Consider this information about 120 students:

	Studies mathematics	Does not study mathematics	Total
Studies physics	36	14	50
Does not study physics	24	46	70
Total	60	60	120

Example

Using a given group as the denominator

One student is selected at random. Find the probability that the student studies mathematics, given that they study physics.

The condition “given that they study physics” restricts attention to the 50 students in the physics row.
Of those 50 students, 36 also study mathematics. These students belong to the intersection of the two events.
Therefore,
$P(mathematics∣physics)=3650=1825.P(\text{mathematics}\mid\text{physics}) =\frac{36}{50} =\frac{18}{25}.$

If the question instead asked for the probability that a student studies physics, given that they study mathematics, the denominator would be 60:

P(physics∣mathematics)=3660=35.P(\text{physics}\mid\text{mathematics}) =\frac{36}{60} =\frac{3}{5}.

These probabilities are different because they use different restricted groups.

Common Mistake

Reversing the condition

In general, $P(A∣B)≠P(B∣A)P(A\mid B)\neq P(B\mid A)$ . The event after the vertical line determines the denominator.

Conditional probability with combined experiments

For experiments involving several stages, it can help to list all equally likely outcomes before applying the condition.

Example

Using ordered outcomes from two dice

Two fair six-sided dice are rolled. Find the probability that both dice show even numbers, given that their total is 8.

Restrict the ordered outcomes to those whose total is 8:
$(6,2).(2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2).$
Among these five equally likely outcomes, both scores are even for
$(6,2).(2,6),\ (4,4),\ (6,2).$
Three of the five outcomes in the restricted sample space satisfy the required event, so the conditional probability is
$35.\frac{3}{5}.$

The outcomes must be treated as ordered pairs: for example, $(2, 6)$ and $(6, 2)$ are different outcomes because the first and second dice have exchanged scores.

Rearranging the conditional probability formula

Starting from

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

multiplying by $P (B)$ gives the multiplication rule

P(A∩B)=P(B)P(A∣B).P(A\cap B)=P(B)P(A\mid B).

Similarly,

P(A∩B)=P(A)P(B∣A).P(A\cap B)=P(A)P(B\mid A).

Therefore,

P(B)P(A∣B)=P(A)P(B∣A).P(B)P(A\mid B)=P(A)P(B\mid A).

This is useful when the conditional probability and the probability of the condition are known.

Example

Finding an intersection probability

Suppose $P (B) = 0.4$ and $P(A∣B)=0.7P(A\mid B)=0.7$ . Find $P(A∩B)P(A\cap B)$ .

Use the multiplication rule because the intersection probability is required:
$P(A∩B)=P(B)P(A∣B).P(A\cap B)=P(B)P(A\mid B).$
Substitute the given probabilities:
$P(A∩B)=0.4×0.7.P(A\cap B)=0.4\times 0.7.$
Hence,
$P(A∩B)=0.28.P(A\cap B)=0.28.$

Conditional probability and independence

Two events are independent if knowing that one has occurred does not change the probability of the other.

Therefore, if $A$ and $B$ are independent and their probabilities are non-zero,

P(A∣B)=P(A)P(A\mid B)=P(A)

and

P(B∣A)=P(B).P(B\mid A)=P(B).

This is equivalent to the familiar independence condition

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

Key Idea

Testing for independence

If conditioning on $B$ changes the probability of $A$ , then $A$ and $B$ are not independent.

Example

Checking whether two events are independent

Suppose $P (A) = 0.3$ , $P (B) = 0.5$ and $P(A∩B)=0.2P(A\cap B)=0.2$ . Determine whether $A$ and $B$ are independent.

Calculate the conditional probability:
$P(A∣B)=P(A∩B)P(B)=0.20.5=0.4.P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{0.2}{0.5} =0.4.$
Compare this with the unconditional probability $P (A) = 0.3$ .
Since $P(A∣B)≠P(A)P(A\mid B)\neq P(A)$ , knowing that $B$ occurred changes the probability of $A$ . The events are therefore not independent.

Tip

Read the notation aloud

Read $P(A∣B)P(A\mid B)$ as “probability of $A$ , given $B$ ”. This helps you identify $B$ as the restricted sample space and prevents you from reversing the condition.

Exam technique

In the exam

Translate “given that” into conditional notation and identify the event after the vertical line.
Restrict the sample space to that condition before counting outcomes or choosing a denominator.
Use $P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ , and check that your answer lies between 0 and 1.
Keep $P(A∣B)P(A\mid B)$ and $P(B∣A)P(B\mid A)$ separate: changing their order usually changes the answer.

Self review

Check yourself

If $P(A∩B)=0.15P(A\cap B)=0.15$ and $P (B) = 0.6$ , what is $P(A∣B)P(A\mid B)$ ?
When a card is known to be a face card, what becomes the new sample space?
How could you use conditional probability to decide whether two events are independent?

What you'll learn

What it means for a probability to be conditional on an event.
How to restrict a sample space when new information is given.
How to calculate conditional probabilities directly from outcomes, frequencies and set information.
How to distinguish conditional probability from intersection and independence.

Prerequisites: events and intersections

A sample space is the set of all possible outcomes of an experiment. An event is a set of outcomes with a particular property.

For example, when rolling a fair six-sided die, the sample space is

S=\{1,2,3,4,5,6\}.

If event $A$ is “the result is even”, then

A=\{2,4,6\}.

If $A$ and $B$ are two events, the intersection $A∩BA\cap B$ contains the outcomes that belong to both events.

For instance, if $B$ is “the result is greater than 3”, then

B=\{4,5,6\}

and

A∩B={4,6}.A\cap B=\{4,6\}.

Definition

Intersection of events

The event $A∩BA\cap B$ occurs when both $A$ and $B$ occur.

What is conditional probability?

Sometimes you are given extra information about the outcome of an experiment. This information changes which outcomes are still possible.

The conditional probability of $A$ given $B$ is the probability that $A$ occurs when you already know that $B$ has occurred. It is written as $P(A∣B)P(A\mid B)$ .

The vertical line means “given that”, so you read $P(A∣B)P(A\mid B)$ as:

the probability of $A$ , given that $B$ has occurred.

Definition

Conditional probability

Provided $P (B) > 0$ , the conditional probability of $A$ given $B$ is

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

The event after the vertical line is the condition. It becomes your new, restricted sample space.

A Venn diagram showing the original sample space and how conditioning on B restricts attention to B, with the intersection of A and B highlighted

Key Idea

Restrict, then compare

To find $P(A∣B)P(A\mid B)$ , restrict attention to outcomes in $B$ , then find the proportion of those outcomes that also belong to $A$ .

Restricting a die sample space

Example

Conditioning on an even die score

A fair die is rolled. Let $A$ be the event “the score is greater than 3” and let $B$ be the event “the score is even”. Find $P(A∣B)P(A\mid B)$ .

Knowing that $B$ occurred restricts the possible outcomes to
$B=\{2,4,6\}.$
Within this restricted sample space, the outcomes greater than 3 are 4 and 6. Therefore,
$A∩B={4,6}.A\cap B=\{4,6\}.$
Two of the three equally likely outcomes in $B$ also belong to $A$ , so
$P(A∣B)=23.P(A\mid B)=\frac{2}{3}.$

Notice that the original sample space had six outcomes, but the denominator is now 3 because the condition tells you the result is even.

Common Mistake

Using the original denominator

Once you are told that $B$ has occurred, outcomes outside $B$ are no longer possible. The denominator must represent the outcomes in $B$ , not the whole original sample space.

Deriving the formula from first principles

Suppose a sample space contains $n (S)$ equally likely outcomes, where $n (X)$ means the number of outcomes in set $X$ .

After conditioning on $B$ , there are $n (B)$ outcomes under consideration. Of these, $n(A∩B)n(A\cap B)$ also satisfy $A$ . Therefore,

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

For equally likely outcomes,

P(A∩B)=n(A∩B)n(S)P(A\cap B)=\frac{n(A\cap B)}{n(S)}

and

P(B)=n(B)n(S).P(B)=\frac{n(B)}{n(S)}.

Dividing these probabilities gives

P(A∩B)P(B)=n(A∩B)n(S)n(B)n(S)=n(A∩B)n(B).\begin{aligned} \frac{P(A\cap B)}{P(B)} &=\frac{\frac{n(A\cap B)}{n(S)}}{\frac{n(B)}{n(S)}}\\ &=\frac{n(A\cap B)}{n(B)}. \end{aligned}

This produces the conditional probability formula directly.

Common Mistake

The condition must be possible

The expression $P(A∣B)P(A\mid B)$ is not defined when $P (B) = 0$ , because the formula would require division by zero.

Calculating from a frequency table

In real contexts, you may be given observed frequencies rather than theoretical probabilities. The same principle applies:

Select the row, column or group named after “given that”.
Use the total frequency of that group as the denominator.
Count how many members of that group also satisfy the required event.

Consider this information about 120 students:

	Studies mathematics	Does not study mathematics	Total
Studies physics	36	14	50
Does not study physics	24	46	70
Total	60	60	120

Example

Using a given group as the denominator

One student is selected at random. Find the probability that the student studies mathematics, given that they study physics.

The condition “given that they study physics” restricts attention to the 50 students in the physics row.
Of those 50 students, 36 also study mathematics. These students belong to the intersection of the two events.
Therefore,
$P(mathematics∣physics)=3650=1825.P(\text{mathematics}\mid\text{physics}) =\frac{36}{50} =\frac{18}{25}.$

If the question instead asked for the probability that a student studies physics, given that they study mathematics, the denominator would be 60:

P(physics∣mathematics)=3660=35.P(\text{physics}\mid\text{mathematics}) =\frac{36}{60} =\frac{3}{5}.

These probabilities are different because they use different restricted groups.

Common Mistake

Reversing the condition

In general, $P(A∣B)≠P(B∣A)P(A\mid B)\neq P(B\mid A)$ . The event after the vertical line determines the denominator.

Conditional probability with combined experiments

For experiments involving several stages, it can help to list all equally likely outcomes before applying the condition.

Example

Using ordered outcomes from two dice

Two fair six-sided dice are rolled. Find the probability that both dice show even numbers, given that their total is 8.

Restrict the ordered outcomes to those whose total is 8:
$(6,2).(2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2).$
Among these five equally likely outcomes, both scores are even for
$(6,2).(2,6),\ (4,4),\ (6,2).$
Three of the five outcomes in the restricted sample space satisfy the required event, so the conditional probability is
$35.\frac{3}{5}.$

The outcomes must be treated as ordered pairs: for example, $(2, 6)$ and $(6, 2)$ are different outcomes because the first and second dice have exchanged scores.

Rearranging the conditional probability formula

Starting from

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

multiplying by $P (B)$ gives the multiplication rule

P(A∩B)=P(B)P(A∣B).P(A\cap B)=P(B)P(A\mid B).

Similarly,

P(A∩B)=P(A)P(B∣A).P(A\cap B)=P(A)P(B\mid A).

Therefore,

P(B)P(A∣B)=P(A)P(B∣A).P(B)P(A\mid B)=P(A)P(B\mid A).

This is useful when the conditional probability and the probability of the condition are known.

Example

Finding an intersection probability

Suppose $P (B) = 0.4$ and $P(A∣B)=0.7P(A\mid B)=0.7$ . Find $P(A∩B)P(A\cap B)$ .

Use the multiplication rule because the intersection probability is required:
$P(A∩B)=P(B)P(A∣B).P(A\cap B)=P(B)P(A\mid B).$
Substitute the given probabilities:
$P(A∩B)=0.4×0.7.P(A\cap B)=0.4\times 0.7.$
Hence,
$P(A∩B)=0.28.P(A\cap B)=0.28.$

Conditional probability and independence

Two events are independent if knowing that one has occurred does not change the probability of the other.

Therefore, if $A$ and $B$ are independent and their probabilities are non-zero,

P(A∣B)=P(A)P(A\mid B)=P(A)

and

P(B∣A)=P(B).P(B\mid A)=P(B).

This is equivalent to the familiar independence condition

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

Key Idea

Testing for independence

If conditioning on $B$ changes the probability of $A$ , then $A$ and $B$ are not independent.

Example

Checking whether two events are independent

Suppose $P (A) = 0.3$ , $P (B) = 0.5$ and $P(A∩B)=0.2P(A\cap B)=0.2$ . Determine whether $A$ and $B$ are independent.

Calculate the conditional probability:
$P(A∣B)=P(A∩B)P(B)=0.20.5=0.4.P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{0.2}{0.5} =0.4.$
Compare this with the unconditional probability $P (A) = 0.3$ .
Since $P(A∣B)≠P(A)P(A\mid B)\neq P(A)$ , knowing that $B$ occurred changes the probability of $A$ . The events are therefore not independent.

Tip

Read the notation aloud

Read $P(A∣B)P(A\mid B)$ as “probability of $A$ , given $B$ ”. This helps you identify $B$ as the restricted sample space and prevents you from reversing the condition.

Exam technique

In the exam

Translate “given that” into conditional notation and identify the event after the vertical line.
Restrict the sample space to that condition before counting outcomes or choosing a denominator.
Use $P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ , and check that your answer lies between 0 and 1.
Keep $P(A∣B)P(A\mid B)$ and $P(B∣A)P(B\mid A)$ separate: changing their order usually changes the answer.

Self review

Check yourself

If $P(A∩B)=0.15P(A\cap B)=0.15$ and $P (B) = 0.6$ , what is $P(A∣B)P(A\mid B)$ ?
When a card is known to be a face card, what becomes the new sample space?
How could you use conditional probability to decide whether two events are independent?

What you'll learn

Prerequisites: events and intersections

What is conditional probability?

Restricting a die sample space

Deriving the formula from first principles

Calculating from a frequency table

Conditional probability with combined experiments

Rearranging the conditional probability formula

Conditional probability and independence

Review 2.3.4 Conditional probability from first principles (A-level only) by teaching Genie

2.3.4 Conditional probability from first principles (A-level only)

What you'll learn

Prerequisites: events and intersections

What is conditional probability?

Restricting a die sample space

Deriving the formula from first principles

Calculating from a frequency table

Conditional probability with combined experiments

Rearranging the conditional probability formula

Conditional probability and independence

Review 2.3.4 Conditional probability from first principles (A-level only) by teaching Genie

2.3.4 Conditional probability from first principles (A-level only)

What you'll learn

Prerequisites: events and intersections

What is conditional probability?

Restricting a die sample space

Deriving the formula from first principles

Calculating from a frequency table

Conditional probability with combined experiments

Rearranging the conditional probability formula

Conditional probability and independence

Review 2.3.4 Conditional probability from first principles (A-level only) by teaching Genie

2.3.4 Conditional probability from first principles (A-level only) Revision Guide

2.3.4 Conditional probability from first principles (A-level only)

What you'll learn

Prerequisites: events and intersections

What is conditional probability?

Restricting a die sample space

Deriving the formula from first principles

Calculating from a frequency table

Conditional probability with combined experiments

Rearranging the conditional probability formula

Conditional probability and independence

Review 2.3.4 Conditional probability from first principles (A-level only) by teaching Genie

2.3.4 Conditional probability from first principles (A-level only) Revision Guide