P9ProbabilityHigherGrades 7–9

Conditional Probability

Q: What does the vertical bar mean?

P(A B) is read "the probability of A given B".

Q: How do I know if events are independent?

If P(A B) = P(A), knowing B does not change the chance of A.

The probability a student plays football is one thing; the probability a student plays football $\textit{given that}$ they are a boy is another. Conditional probability asks about one event when you already know another has happened, and it appears in AQA questions with two-way tables, Venn diagrams and tree diagrams.

45 min AQA GCSE Maths

The big picture

Knowing that an event has happened shrinks the set of possibilities. For "the probability of $A$ given $B$ ", written $P(A \mid B)$ , you only look at outcomes where $B$ happened, and ask what fraction of those are also $A$ .

With counts in a table or Venn diagram, this is just a fraction with a smaller denominator. On a tree diagram, the second set of branches already are conditional probabilities — which is why picking without replacement changes them.

What you'll learn

Find conditional probabilities from two-way tables
Find conditional probabilities from Venn diagrams
Use tree diagrams for dependent events
Work backwards to find "given that" probabilities from a tree

From a two-way table

80 students were asked whether they play football. Boys: 27 yes, 13 no. Girls: 18 yes, 22 no.

Given that a student plays football, the denominator is the 45 who play: $P(\text{boy} \mid \text{football}) = \dfrac{27}{45} = 0.6$ .

Given that the student is a girl, the denominator is 40: $P(\text{football} \mid \text{girl}) = \dfrac{18}{40} = 0.45$ .

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

In a class, 30 boys and 20 girls were asked if they like maths. 18 boys and 12 girls said yes. A student who likes maths is chosen at random. What is the probability that the student is a girl?

Your answer

Show worked solution

Solution

1Students who like maths: $18 + 12 = 30$ .
2Of those, 12 are girls: $\frac{12}{30} = \frac{2}{5}$ .

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

Tip — Underline the "given that" condition. It tells you which row or column total to divide by.

From a Venn diagram

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ , or with counts, $\dfrac{n(A \cap B)}{n(B)}$ .

Example 1In a group of 50, 12 are in

A

only, 8 are in both

A

and

B

, 20 are in

B

only, and 10 are in neither. Find

P(A \mid B)

n(B) = 8 + 20 = 28

n(A \cap B) = 8

Answer

P(A \mid B) = \dfrac{8}{28} = \dfrac27

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

In the group of 50 (12 in $A$ only, 8 in both, 20 in $B$ only, 10 in neither), find $P(B \mid A)$ .

Your answer

Show worked solution

Solution

1Inside $A$ : $12 + 8 = 20$ people.
2Of those, 8 are also in $B$ : $\frac{8}{20} = \frac{2}{5}$ .

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

Tree diagrams for dependent events

Events are dependent when the first changes the probabilities for the second — for example, taking counters from a bag without replacing them.

A bag holds 4 red and 6 blue counters. Two are taken without replacement. After a red, 3 of the 9 remaining are red, so the second-branch probability is $\dfrac39$ .

Example 1Find the probability both counters are red, and the probability at least one is red.

P(RR) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{2}{15}

P(BB) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac13

, so

P(\text{at least one red}) = 1 - \dfrac13 = \dfrac23

Answer

P(RR) = \dfrac{2}{15}

P(\text{at least one red}) = \dfrac23

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

A bag has 4 red and 6 blue counters. Two are taken without replacement. Given that the first is red, what is the probability that the second is red?

Your answer

Show worked solution

Solution

1After one red is removed, 9 counters remain and 3 are red.
2 $P = \frac{3}{9} = \frac{1}{3}$ .

Answer: $\frac{1}{3}$

Why this works

Every probability on a second branch is conditional: it is the probability of that outcome given what happened on the first branch.

Working backwards

To find a "given that" probability from a tree, divide the probability of the path you want by the total probability of the condition.

Using the counters: $P(\text{both red} \mid \text{at least one red}) = \dfrac{2/15}{2/3} = \dfrac15$ .

Tip — Write the condition’s total probability first, then pick out which of its paths also satisfy the question.

Think like an examiner

Show the restricted total you are dividing by.
Label tree branches with probabilities that reflect any change.
Leave answers as simplified fractions where possible.

Remember these

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

Conditional probability.

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

Multiplying along tree branches.

P(A \mid B) = P(A) \iff \text{independent}

Independence test.

Watch out for these

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

Divide by the total for the condition.

Keeping the same second-branch probabilities without replacement.

Reduce the numbers after each pick.

Confusing

P(A \mid B)

with

P(B \mid A)

The event after the bar is the one known to have happened.

Adding along branches.

Multiply along branches, add between paths.

Stretch yourself

The probability Kai passes a driving test at the first attempt is 0.6. If Kai fails, the probability of passing at the second attempt is 0.75. (a) Find the probability Kai passes within two attempts. (b) Given that Kai passes within two attempts, find the probability Kai failed the first attempt.

Hint — There are two paths that end in a pass.

Reveal solution

Solution(a) Pass first: 0.6. Fail then pass:

0.4 \times 0.75 = 0.3

. Total

= 0.9

. (b)

P(\text{failed first} \mid \text{passed within two}) = \dfrac{0.3}{0.9} = \dfrac13

\blacksquare

Questions students ask

What does the vertical bar mean?

P(A \mid B)

is read "the probability of A given B".

How do I know if events are independent?

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

, knowing B does not change the chance of A.

Should I use a table or a tree?

Use whatever the question gives; if it gives neither, a tree is usually quickest for sequences of events.

Key takeaways

"Given that" restricts the outcomes you count.
Divide by the total of the known condition.
Tree branches after the first are conditional probabilities.
Without replacement, update the numbers on later branches.
Work backwards by dividing a path by the condition’s total.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 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 Year 10, 27 of 45 students have a phone case. In Year 11, 28 of 35 students have a phone case.

A student with a phone case is chosen at random. Work out the probability that they are in Year 11.[2]
Show mark scheme
- M1Total with a case: $27 + 28 = 55$
- A1 $\frac{28}{55}$
Answer: $\frac{28}{55}$

Question 2Grade 84 marks

A bag contains 5 red and 3 green sweets. Two sweets are taken at random without replacement.

Given that at least one of the sweets is red, work out the probability that both are red.[4]
Show mark scheme
- M1 $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$
- M1 $P(\text{at least one red}) = 1 - \frac{3}{8} \times \frac{2}{7} = \frac{50}{56}$
- M1Divides: $\frac{20}{56} \div \frac{50}{56}$
- A1 $\frac{2}{5}$
Answer: $\frac{2}{5}$

Question 3Grade 93 marks

$P(A) = 0.4$ . $P(B \mid A) = 0.3$ and $P(B \mid A') = 0.6$ .

Work out $P(A \mid B)$ .[3]
Show mark scheme
- M1 $P(A \cap B) = 0.4 \times 0.3 = 0.12$
- M1 $P(B) = 0.12 + 0.6 \times 0.6 = 0.48$
- A1 $\frac{0.12}{0.48} = 0.25$
Answer: $0.25$

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