1MA1 · P9ProbabilityHigherGrades 7–9

Conditional Probability

Q: What does the vertical line mean?

"Given that". P(A B) is the probability of A given that B has happened.

Q: How does conditional probability relate to independence?

If A and B are independent, knowing B does not change the probability of A, so P(A B) = P(A).

The probability that a randomly chosen student plays tennis is one thing. The probability that a student plays tennis $\textit{given that}$ they play football is another — you already know something about them, so you only look at the footballers. That is $\textbf{conditional probability}$ , and it is the most challenging probability idea at GCSE.

50 min Probability

The big picture

"Given that" shrinks the sample space. Instead of dividing by everyone, divide by the group you are told about. In a Venn diagram or two-way table, that means using the total of the given set as the denominator.

Written as P( $A \mid B$ ), the probability of $A$ given $B$ equals $\dfrac{P(A \cap B)}{P(B)}$ . Edexcel sets conditional probability questions using Venn diagrams, two-way tables and tree diagrams, and they are usually worth 2–4 marks each.

What you'll learn

Understand and use the notation P( $A \mid B$ )
Calculate conditional probabilities from two-way tables and Venn diagrams
Use tree diagrams with dependent events
Work backwards to find a conditional probability from a tree diagram (Higher)

What "given that" means

P( $A \mid B$ ) is read "the probability of $A$ given $B$ ". You already know $B$ has happened, so only outcomes in $B$ are possible.

So P( $A \mid B$ ) $= \dfrac{\text{number in both } A \text{ and } B}{\text{number in } B}$ .

Using probabilities: P( $A \mid B$ ) $= \dfrac{P(A \cap B)}{P(B)}$ .

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}

Conditional probability.

Example 1In a class of 30, 18 play football, 12 play tennis and 5 play both. A footballer is chosen at random. Find the probability they also play tennis.

1Given football: only the 18 footballers are possible.

2Of those, 5 play tennis.

AnswerP(tennis | football)

= \tfrac{5}{18}

Your turnSame method, new numbers — try it before moving on.

In a class of 40, 22 dance, 15 sing and 6 do both. A singer is chosen at random. Find the probability that they also dance.

Your answer

Show worked solution

Solution

1Only the 15 singers count.
26 of them dance: $\frac{6}{15} = \frac{2}{5}$ .

Answer: $\frac{2}{5}$

Tip — Underline the "given" condition in the question. Its total is your denominator.

Two-way tables

A two-way table shows counts for two categories. For a conditional probability, find the row or column for the given condition and use its total.

Example 180 students choose a trip. Of the 45 in Year 10, 27 choose the museum; of the 35 in Year 11, 14 choose the museum. A student who chose the museum is picked at random. Find the probability they are in Year 11.

1Museum total:

27 + 14 = 41

2Year 11 among those: 14.

Answer

\tfrac{14}{41}

Your turnSame method, new numbers — try it before moving on.

Of 30 adults, 18 own a dog. Of 20 children, 8 own a dog. A dog owner is chosen at random. Find the probability it is a child.

Your answer

Show worked solution

Solution

1Dog owners: $18 + 8 = 26$ .
2Children among them: $\frac{8}{26} = \frac{4}{13}$ .

Answer: $\frac{4}{13}$

Why this works

P(Year 11 | museum) $= \tfrac{14}{41}$ is not the same as P(museum | Year 11) $= \tfrac{14}{35}$ . Swapping the condition changes the denominator — confusing the two is the classic error.

Dependent events on trees

On a tree diagram, second-stage branches often show conditional probabilities: P(second event | first outcome).

Multiplying along a path uses P( $A$ and $B$ ) $=$ P( $A$ ) × P( $B \mid A$ ).

Example 1The probability it rains is 0.3. If it rains, the probability the bus is late is 0.6; if not, it is 0.2. Find the probability the bus is late.

1Rain and late:

0.3 \times 0.6 = 0.18

2No rain and late:

0.7 \times 0.2 = 0.14

3Add the paths.

Answer

0.32

Your turnSame method, new numbers — try it before moving on.

A team wins its first game with probability 0.4. If it wins, it wins the second with probability 0.7; if it loses, it wins the second with probability 0.2. Find the probability it wins the second game.

Your answer

Show worked solution

Solution

1Win then win: $0.4 \times 0.7 = 0.28$ .
2Lose then win: $0.6 \times 0.2 = 0.12$ . Total $0.4$ .

Answer: $0.4$

Working backwards

Sometimes you know the second event happened and want the probability of the first: for example, given the bus was late, what is the probability it rained?

Use P(rain | late) $= \dfrac{P(\text{rain and late})}{P(\text{late})}$ , taking both from the tree.

Example 1Using the bus example, find the probability it rained, given that the bus was late.

1P(rain and late)

= 0.18

2P(late)

= 0.32

\dfrac{0.18}{0.32}

Answer

0.5625

Your turnSame method, new numbers — try it before moving on.

Using the same team, find the probability it won the first game, given that it won the second.

Your answer

Show worked solution

Solution

1 $P(\text{won first and second}) = 0.28$ .
2Divide by $P(\text{won second}) = 0.4$ : $0.7$ .

Answer: $0.7$

Tip — The numerator is one path; the denominator is the sum of all paths that satisfy the given condition.

Think like an examiner

Identify the given condition and use its total as the denominator.
On trees, show the paths used in both numerator and denominator.
Leave answers as fractions or exact decimals unless told to round.

Conditional probability

P(A \mid B) = \tfrac{P(A \cap B)}{P(B)}

Definition.

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

Multiplying along a tree.

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

If A and B are independent.

Watch out for these

Dividing by the grand total for a "given that" probability.

Divide by the total of the given group.

Confusing P(

A \mid B

) with P(

B \mid A

The event after the bar is the condition.

Using only one path for the denominator when working backwards.

Add every path that satisfies the condition.

Treating dependent events as independent.

Use the conditional probabilities on the second branches.

Stretch yourself

A test for a condition that affects 1 in 50 people gives a positive result for 95% of people with the condition and for 4% of people without it. A person tests positive. Find the probability that they have the condition, to 3 significant figures.

Hint — Draw a tree: condition or not, then positive or negative. Divide the "condition and positive" path by all positive paths.

Reveal solution

SolutionP(condition and positive)

= 0.02 \times 0.95 = 0.019

. P(no condition and positive)

= 0.98 \times 0.04 = 0.0392

. P(positive)

= 0.0582

. So P(condition | positive)

= \dfrac{0.019}{0.0582} = 0.326

. Despite the test being quite accurate, only about a third of positive results are genuine, because the condition is rare.

\blacksquare

Questions students ask

What does the vertical line mean?

"Given that". P(

A \mid B

) is the probability of

A

given that

B

has happened.

How does conditional probability relate to independence?

A

and

B

are independent, knowing

B

does not change the probability of

A

, so P(

A \mid B

)

=

A

Where does conditional probability appear in real life?

Medical testing, weather forecasts and insurance all depend on probabilities given certain information.

Key takeaways

"Given that" restricts outcomes to the given group.
P( $A \mid B$ ) $= \tfrac{P(A \cap B)}{P(B)}$ , or count in both ÷ count in the given group.
On trees, second branches are conditional probabilities.
Working backwards: one path ÷ sum of paths meeting the condition.
P( $A \mid B$ ) and P( $B \mid A$ ) are usually different.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

In a group of 60 people, 34 have been to America, 20 have been to Brazil and 8 have been to both.

A person who has been to America is chosen at random. Work out the probability they have also been to Brazil.[2]
Show mark scheme
- M1Denominator 34
- A1 $\frac{8}{34} = \frac{4}{17}$
Answer: $\frac{4}{17}$

Question 2Grade 83 marks

Ana travels by car with probability 0.6, otherwise by bus. By car she is late with probability 0.1; by bus, with probability 0.25.

Given that Ana is late, work out the probability she travelled by bus.[3]
Show mark scheme
- M1 $P(\text{late}) = 0.06 + 0.1 = 0.16$
- M1 $\frac{0.1}{0.16}$
- A1 $0.625$
Answer: $0.625$

Question 3Grade 93 marks

A bag contains 6 red and 4 blue counters. Two counters are taken at random without replacement.

Given that the second counter is red, work out the probability that the first counter was red.[3]
Show mark scheme
- M1 $P(RR) = \frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$
- M1 $P(\text{second red}) = \frac{1}{3} + \frac{4}{10} \times \frac{6}{9} = \frac{3}{5}$
- A1 $\frac{5}{9}$
Answer: $\frac{5}{9}$

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