2.3.3 Conditional probability (A-level only)

What you'll learn

What conditional probability means and how to calculate it.
How to use tree diagrams, Venn diagrams and two-way tables.
How conditional probability connects with intersections and independence.
How to choose an efficient method for a given problem.

Probability language you need

An event is a set of possible outcomes. For example, when rolling a fair die, the event “even number” is the set ${2,4,6\}$ .

The sample space is the set of all possible outcomes.

If $A$ and $B$ are events, their intersection, written $A∩BA\cap B$ , is the event that both $A$ and $B$ occur.

Their union, written $A∪BA\cup B$ , is the event that at least one of $A$ or $B$ occurs, including the possibility that both occur.

For any two events,

P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

The intersection is subtracted because outcomes in both events were counted twice.

What is conditional probability?

Sometimes you are given extra information that reduces the possible outcomes. A conditional probability is a probability calculated using this restricted sample space.

The notation $P(A∣B)P(A\mid B)$ means:

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

The vertical line is read as “given”.

Definition

Conditional probability

Provided $P (B) > 0$ ,

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

The denominator is the probability of the event you are given, while the numerator is the probability that both events occur.

Conditioning on $B$ means that you ignore everything outside $B$ . Within this smaller sample space, the favourable region for $A$ is $A∩BA\cap B$ .

Venn diagram showing that conditioning on B restricts the sample space to B, with the intersection of A and B as the favourable region

Example

Finding a conditional probability from event probabilities

Suppose $P (A) = 0.6$ , $P (B) = 0.5$ and $P(A∩B)=0.3P(A\cap B)=0.3$ . Find $P(A∣B)P(A\mid B)$ .

Since the condition is $B$ , use $P (B)$ as the denominator:
$P(A∣B)=P(A∩B)P(B).P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$
Substitute the given probabilities:
$P(A∣B)=0.30.5.P(A\mid B)=\frac{0.3}{0.5}.$
Simplify to obtain:
$P(A∣B)=0.6.P(A\mid B)=0.6.$

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 restricted sample space and therefore the denominator.

The multiplication rule

Rearranging the conditional probability formula gives the multiplication rule:

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

Equally, by conditioning the other way,

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

These expressions calculate the same intersection, so you can use whichever contains the information available.

Key Idea

Along means multiply

When events happen along one complete route, multiply their probabilities. In symbols, the probability that $A$ occurs and then $B$ occurs is $P(A)P(B∣A)P(A)P(B\mid A)$ .

Example

Using the multiplication rule

A college reports that 55% of its students study mathematics. Of the students who study mathematics, 40% also study physics. Find the probability that a randomly selected student studies both subjects.

Let $M$ be the event “studies mathematics” and $F$ be the event “studies physics”. The information gives $P (M) = 0.55$ and $P(F∣M)=0.40P(F\mid M)=0.40$ .
Apply the multiplication rule:
$P(M∩F)=P(M)P(F∣M).P(M\cap F)=P(M)P(F\mid M).$
Calculate:
$P(M∩F)=0.55×0.40=0.22.P(M\cap F)=0.55\times 0.40=0.22.$
Therefore, the probability is 0.22, or 22%.

Conditional probability in tree diagrams

A tree diagram displays probabilities through a sequence of stages.

Each set of branches leaving one point must have probabilities that add to 1. To find the probability of a complete route, multiply along its branches. To combine mutually exclusive routes leading to the same required event, add their probabilities.

When sampling without replacement, the first outcome changes what remains, so the probabilities on the second set of branches are conditional probabilities.

The diagram below shows two counters drawn without replacement from a bag containing three red and two blue counters.

Probability tree for drawing two counters without replacement from three red and two blue counters

Example

Using a tree diagram without replacement

Two counters are drawn without replacement from a bag containing three red and two blue counters. Find the probability that the counters are different colours.

There are two possible routes to different colours: red then blue, or blue then red.
Multiply along each route:
$B)=35×24=310,P(R\text{ then }B)=\frac35\times\frac24=\frac3{10},$ $R)=25×34=310.P(B\text{ then }R)=\frac25\times\frac34=\frac3{10}.$
The routes are mutually exclusive, so add their probabilities:
$colours)=310+310=35.P(\text{different colours})=\frac3{10}+\frac3{10}=\frac35.$

Tip

Updating probabilities without replacement

After an object is removed, reduce the total number of objects by one and also reduce the relevant category count. Do this before writing the next branch probabilities.

Conditional probability in Venn diagrams

A Venn diagram is useful when events overlap. The intersection $A∩BA\cap B$ lies inside both circles.

To find $P(A∣B)P(A\mid B)$ , treat the whole of $B$ as the new sample space. Compare the part of $B$ that also lies in $A$ with the total size of $B$ .

If a diagram contains frequencies rather than probabilities, the same idea gives

B.P(A\mid B)=\frac{\text{frequency in }A\cap B}{\text{frequency in }B}.

Example

Using frequencies in a Venn diagram

In a group of students, 18 study French, and 10 study both French and Spanish. One student is chosen from those who study French. Find the probability that the student also studies Spanish.

The phrase “chosen from those who study French” makes French the condition, so the denominator is 18.
Of those 18 students, 10 are in the intersection and also study Spanish.
Therefore,
$P(Spanish∣French)=1018=59.P(\text{Spanish}\mid\text{French})=\frac{10}{18}=\frac59.$

Conditional probability in two-way tables

A two-way table organises frequencies using two categories. Row and column totals help you identify the restricted group.

Suppose a survey gives the following results:

	Cycles	Does not cycle	Total
Student	24	36	60
Staff	16	24	40
Total	40	60	100

Example

Using a two-way table

One person is selected from the staff. Find the probability that this person cycles.

“Selected from the staff” restricts the sample space to the staff row, which contains 40 people.
The relevant intersection is “staff and cycles”, containing 16 people.
Hence,
$P(cycles∣staff)=1640=25.P(\text{cycles}\mid\text{staff})=\frac{16}{40}=\frac25.$

Common Mistake

Using the grand total

For a conditional probability, do not automatically divide by the grand total. Divide by the total of the group named after “given” or implied by “selected from”.

Conditional probability and independence

Two events are independent if the occurrence of one does not change the probability of the other.

Provided the conditioning event has non-zero probability, independence means

P(A∣B)=P(A).P(A\mid B)=P(A).

Equivalent tests are

P(B∣A)=P(B)P(B\mid A)=P(B)

and

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

If these equalities do not hold, the events are dependent.

Example

Testing whether events are independent

Suppose $P (A) = 0.4$ , $P (B) = 0.7$ and $P(A∩B)=0.28P(A\cap B)=0.28$ . Determine whether $A$ and $B$ are independent.

Calculate the product of the individual probabilities:
$P(A)P(B)=0.4×0.7=0.28.P(A)P(B)=0.4\times0.7=0.28.$
Compare this with the given intersection probability. Since
$P(A∩B)=P(A)P(B),P(A\cap B)=P(A)P(B),$
the independence condition is satisfied.
Therefore, $A$ and $B$ are independent.

Common Mistake

Mutually exclusive is not independent

If two events with non-zero probabilities are mutually exclusive, one occurring prevents the other from occurring. They are therefore dependent, not independent.

Exam technique

In the exam

Translate “given” into a condition and identify the restricted sample space before calculating.
On a tree, multiply along branches and add separate routes; update branch probabilities when sampling without replacement.
In a Venn diagram or two-way table, use the total of the conditioned group as the denominator.
Keep intersections precise: “both” means $A∩BA\cap B$ , while “at least one” means $A∪BA\cup B$ .
Test independence using $P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)$ when all three probabilities are available.

Self review

Check yourself

Can you explain why $P(A∣B)P(A\mid B)$ and $P(B∣A)P(B\mid A)$ usually have different denominators?
How would the second-stage probabilities on a tree change when sampling without replacement?
Which equality could you use to decide whether two events are independent?

What you'll learn

What conditional probability means and how to calculate it.
How to use tree diagrams, Venn diagrams and two-way tables.
How conditional probability connects with intersections and independence.
How to choose an efficient method for a given problem.

Probability language you need

An event is a set of possible outcomes. For example, when rolling a fair die, the event “even number” is the set ${2,4,6\}$ .

The sample space is the set of all possible outcomes.

If $A$ and $B$ are events, their intersection, written $A∩BA\cap B$ , is the event that both $A$ and $B$ occur.

Their union, written $A∪BA\cup B$ , is the event that at least one of $A$ or $B$ occurs, including the possibility that both occur.

For any two events,

P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

The intersection is subtracted because outcomes in both events were counted twice.

What is conditional probability?

Sometimes you are given extra information that reduces the possible outcomes. A conditional probability is a probability calculated using this restricted sample space.

The notation $P(A∣B)P(A\mid B)$ means:

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

The vertical line is read as “given”.

Definition

Conditional probability

Provided $P (B) > 0$ ,

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

The denominator is the probability of the event you are given, while the numerator is the probability that both events occur.

Conditioning on $B$ means that you ignore everything outside $B$ . Within this smaller sample space, the favourable region for $A$ is $A∩BA\cap B$ .

Venn diagram showing that conditioning on B restricts the sample space to B, with the intersection of A and B as the favourable region

Example

Finding a conditional probability from event probabilities

Suppose $P (A) = 0.6$ , $P (B) = 0.5$ and $P(A∩B)=0.3P(A\cap B)=0.3$ . Find $P(A∣B)P(A\mid B)$ .

Since the condition is $B$ , use $P (B)$ as the denominator:
$P(A∣B)=P(A∩B)P(B).P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$
Substitute the given probabilities:
$P(A∣B)=0.30.5.P(A\mid B)=\frac{0.3}{0.5}.$
Simplify to obtain:
$P(A∣B)=0.6.P(A\mid B)=0.6.$

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 restricted sample space and therefore the denominator.

The multiplication rule

Rearranging the conditional probability formula gives the multiplication rule:

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

Equally, by conditioning the other way,

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

These expressions calculate the same intersection, so you can use whichever contains the information available.

Key Idea

Along means multiply

When events happen along one complete route, multiply their probabilities. In symbols, the probability that $A$ occurs and then $B$ occurs is $P(A)P(B∣A)P(A)P(B\mid A)$ .

Example

Using the multiplication rule

Let $M$ be the event “studies mathematics” and $F$ be the event “studies physics”. The information gives $P (M) = 0.55$ and $P(F∣M)=0.40P(F\mid M)=0.40$ .
Apply the multiplication rule:
$P(M∩F)=P(M)P(F∣M).P(M\cap F)=P(M)P(F\mid M).$
Calculate:
$P(M∩F)=0.55×0.40=0.22.P(M\cap F)=0.55\times 0.40=0.22.$
Therefore, the probability is 0.22, or 22%.

Conditional probability in tree diagrams

A tree diagram displays probabilities through a sequence of stages.

When sampling without replacement, the first outcome changes what remains, so the probabilities on the second set of branches are conditional probabilities.

The diagram below shows two counters drawn without replacement from a bag containing three red and two blue counters.

Probability tree for drawing two counters without replacement from three red and two blue counters

Example

Using a tree diagram without replacement

Two counters are drawn without replacement from a bag containing three red and two blue counters. Find the probability that the counters are different colours.

There are two possible routes to different colours: red then blue, or blue then red.
Multiply along each route:
$B)=35×24=310,P(R\text{ then }B)=\frac35\times\frac24=\frac3{10},$ $R)=25×34=310.P(B\text{ then }R)=\frac25\times\frac34=\frac3{10}.$
The routes are mutually exclusive, so add their probabilities:
$colours)=310+310=35.P(\text{different colours})=\frac3{10}+\frac3{10}=\frac35.$

Tip

Updating probabilities without replacement

After an object is removed, reduce the total number of objects by one and also reduce the relevant category count. Do this before writing the next branch probabilities.

Conditional probability in Venn diagrams

A Venn diagram is useful when events overlap. The intersection $A∩BA\cap B$ lies inside both circles.

To find $P(A∣B)P(A\mid B)$ , treat the whole of $B$ as the new sample space. Compare the part of $B$ that also lies in $A$ with the total size of $B$ .

If a diagram contains frequencies rather than probabilities, the same idea gives

B.P(A\mid B)=\frac{\text{frequency in }A\cap B}{\text{frequency in }B}.

Example

Using frequencies in a Venn diagram

In a group of students, 18 study French, and 10 study both French and Spanish. One student is chosen from those who study French. Find the probability that the student also studies Spanish.

The phrase “chosen from those who study French” makes French the condition, so the denominator is 18.
Of those 18 students, 10 are in the intersection and also study Spanish.
Therefore,
$P(Spanish∣French)=1018=59.P(\text{Spanish}\mid\text{French})=\frac{10}{18}=\frac59.$

Conditional probability in two-way tables

A two-way table organises frequencies using two categories. Row and column totals help you identify the restricted group.

Suppose a survey gives the following results:

	Cycles	Does not cycle	Total
Student	24	36	60
Staff	16	24	40
Total	40	60	100

Example

Using a two-way table

One person is selected from the staff. Find the probability that this person cycles.

“Selected from the staff” restricts the sample space to the staff row, which contains 40 people.
The relevant intersection is “staff and cycles”, containing 16 people.
Hence,
$P(cycles∣staff)=1640=25.P(\text{cycles}\mid\text{staff})=\frac{16}{40}=\frac25.$

Common Mistake

Using the grand total

For a conditional probability, do not automatically divide by the grand total. Divide by the total of the group named after “given” or implied by “selected from”.

Conditional probability and independence

Two events are independent if the occurrence of one does not change the probability of the other.

Provided the conditioning event has non-zero probability, independence means

P(A∣B)=P(A).P(A\mid B)=P(A).

Equivalent tests are

P(B∣A)=P(B)P(B\mid A)=P(B)

and

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

If these equalities do not hold, the events are dependent.

Example

Testing whether events are independent

Suppose $P (A) = 0.4$ , $P (B) = 0.7$ and $P(A∩B)=0.28P(A\cap B)=0.28$ . Determine whether $A$ and $B$ are independent.

Calculate the product of the individual probabilities:
$P(A)P(B)=0.4×0.7=0.28.P(A)P(B)=0.4\times0.7=0.28.$
Compare this with the given intersection probability. Since
$P(A∩B)=P(A)P(B),P(A\cap B)=P(A)P(B),$
the independence condition is satisfied.
Therefore, $A$ and $B$ are independent.

Common Mistake

Mutually exclusive is not independent

If two events with non-zero probabilities are mutually exclusive, one occurring prevents the other from occurring. They are therefore dependent, not independent.

Exam technique

In the exam

Translate “given” into a condition and identify the restricted sample space before calculating.
On a tree, multiply along branches and add separate routes; update branch probabilities when sampling without replacement.
In a Venn diagram or two-way table, use the total of the conditioned group as the denominator.
Keep intersections precise: “both” means $A∩BA\cap B$ , while “at least one” means $A∪BA\cup B$ .
Test independence using $P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)$ when all three probabilities are available.

Self review

Check yourself

Can you explain why $P(A∣B)P(A\mid B)$ and $P(B∣A)P(B\mid A)$ usually have different denominators?
How would the second-stage probabilities on a tree change when sampling without replacement?
Which equality could you use to decide whether two events are independent?

What you'll learn

Probability language you need

What is conditional probability?

The multiplication rule

Conditional probability in tree diagrams

Conditional probability in Venn diagrams

Conditional probability in two-way tables

Conditional probability and independence

Review 2.3.3 Conditional probability (A-level only) by teaching Genie

2.3.3 Conditional probability (A-level only)

What you'll learn

Probability language you need

What is conditional probability?

The multiplication rule

Conditional probability in tree diagrams

Conditional probability in Venn diagrams

Conditional probability in two-way tables

Conditional probability and independence

Review 2.3.3 Conditional probability (A-level only) by teaching Genie

2.3.3 Conditional probability (A-level only)

What you'll learn

Probability language you need

What is conditional probability?

The multiplication rule

Conditional probability in tree diagrams

Conditional probability in Venn diagrams

Conditional probability in two-way tables

Conditional probability and independence

Review 2.3.3 Conditional probability (A-level only) by teaching Genie

2.3.3 Conditional probability (A-level only) Revision Guide

2.3.3 Conditional probability (A-level only)

What you'll learn

Probability language you need

What is conditional probability?

The multiplication rule

Conditional probability in tree diagrams

Conditional probability in Venn diagrams

Conditional probability in two-way tables

Conditional probability and independence

Review 2.3.3 Conditional probability (A-level only) by teaching Genie

2.3.3 Conditional probability (A-level only) Revision Guide