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GCSE Maths · Probability
Conditional probability GCSE Questions and Worked Answers
Conditional probability is the chance of an event within a group already known to satisfy a condition. Restrict the possible outcomes to the 'given' group, then divide the favourable count by that group's total. P(A|B) means the probability of A given B.
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Start with the meaning
What you need to know about conditional probability
A class has 40 pupils. Of the 24 pupils who take the bus, 6 are late. If you choose randomly from the whole class, all 40 pupils are possible. But if you are told the chosen pupil takes the bus, only those 24 remain possible. The chance of being late is then 6 out of 24. This is a conditional probability: the chance after using the extra information.
See the idea first
Cover the groups that the condition rules out
In the table, 'given bus' means use only the bus row. The favourable 6 are late bus users; the denominator 24 counts every bus user. Write P(late | bus) = 6/24. The vertical bar means 'given'. In general, P(A | B) = P(A and B)/P(B), provided P(B) is not zero: both probabilities are measured over the whole sample, and the division rescales them to the given group. In this table, (6/40) ÷ (24/40) = 6/24.
| On time | Late | Total | |
|---|---|---|---|
| Bus | 18 | 6 | 24 |
| Walk | 14 | 2 | 16 |
| Total | 32 | 8 | 40 |
From problem to method
Typical problems you need to be able to solve
These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.
Read a table condition
What the problem asks: Using the table, find the chance a pupil walks, given they are late.
How to solve it: Restrict to the late column, whose total is 8. Two of those 8 walk, so the probability is 2/8 = 1/4.
Update after a known draw
What the problem asks: A bag has 5 red and 3 blue counters. One red is removed. Find the chance the next counter is blue.
How to solve it: There are now 7 counters, still including 3 blue. The probability is 3/7. Only update the colours actually removed.
Use probabilities rather than counts
What the problem asks: P(A and B) = 0.12 and P(B) = 0.3. Find P(A | B).
How to solve it: Divide the joint probability by the probability of the condition: 0.12/0.3 = 0.4.
A reliable routine
Find probability within the known condition
Use this when extra information restricts the possible outcomes, especially 'given that'. Dividing by the size or probability of that restricted group re-expresses the favourable part as a fraction of the remaining possibilities.
- State exactly what is known and restrict to that group.
- Within that group, identify the favourable outcomes.
- Divide the favourable count by the group total, or joint probability by condition probability.
- Check the fraction lies between 0 and 1 and that the condition is the denominator.
Check: P(A | B) usually differs from P(B | A). 'Given late' and 'given bus' select different groups even when the favourable overlap is the same.
Fully worked
Conditional probability GCSE worked examples
Each solution explains the clue, why the method fits and how to check the result.
Example 1
A given row
Question
A club has 50 members. Of 30 swimmers, 12 also run. A swimmer is chosen at random. Find the probability they also run.
The condition restricts the choice to 30 swimmers.
The total club size is not the denominator.
Example 2
Reverse the condition
Question
A club has 30 swimmers and 20 runners; 12 do both. Compare P(run | swim) with P(swim | run).
The numerator is the same overlap, but the known group changes.
Example 3
A complement inside the group
Question
Of 28 pupils taking music, 9 also take drama. A music pupil is chosen at random. Find the probability they do not take drama.
Within the music group, 28 − 9 = 19 do not take drama.
The complement is taken inside the music group.
Example 4
Without replacement
Question
A bag contains 6 green and 4 yellow counters. A green counter is removed and not replaced. Find the probability the next counter is green.
There are now 5 green among 9 total.
Both the green count and total fall by one.
Example 5
Joint probability supplied
Question
P(A and B) = 0.18 and P(B) = 0.45. Find P(A | B).
The event after the bar supplies the denominator.
Example 6
A condition on two dice
Question
Two fair six-sided dice are rolled independently. Given that their sum is 8, find the probability that at least one die shows 3.
The equally likely ordered pairs with sum 8 are (2,6), (3,5), (4,4), (5,3), (6,2). Two contain a 3.
We count within the five allowed pairs, not all 36.
10 original questions · total 22 marks
Conditional probability GCSE exam-style questions
Try each question before opening its fully worked answer. The difficulty rises through the set.
Before you startAllow about 27 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
Restricted group
18 of 45 pupils cycle. Of those cyclists, 6 wear a red coat. Find P(red coat | cyclist).
Show worked answer
Only the 18 cyclists are possible.
Reverse group
There are 18 cyclists and 10 red-coat wearers; 6 belong to both groups. Find P(cyclist | red coat).
Show worked answer
The condition selects the 10 red-coat wearers.
Not in an event
Of 25 chess players, 7 also play tennis. Find P(not tennis | chess).
Show worked answer
Known red removed
A bag has 4 red and 5 blue counters. One red is removed without replacement. Find P(next blue).
Show worked answer
There are 8 counters left and still 5 blue.
Known blue removed
A bag has 4 red and 5 blue counters. One blue is removed without replacement. Find P(next blue).
Show worked answer
Now 4 blue remain out of 8.
With replacement
A bag has 4 red and 5 blue counters. A blue is drawn and replaced. Find the chance the next draw is blue.
Show worked answer
Because the counter was put back, the bag has its original 4 red and 5 blue counters again.
The bag is unchanged, so the probability is 5/9 again.
Divide probabilities
P(A and B) = 0.15 and P(B) = 0.6. Find P(A | B).
Show worked answer
Recover the overlap
P(A | B) = 0.4 and P(B) = 0.35. Find P(A and B).
Show worked answer
Multiply by the condition probability:
Dice condition
Two fair six-sided dice are rolled independently. Given a total of 5, find the probability the first die is 2.
Show worked answer
Allowed pairs: (1,4), (2,3), (3,2), (4,1). Only (2,3) has first die 2.
Impossible condition
Why cannot the usual formula define P(A | B) if P(B) = 0?
Show worked answer
It would divide by zero. The condition has no positive-probability group to use as the new denominator, so the formula is undefined.
Examiner-style feedback
Common conditional probability mistakes
The given group, not the original total, supplies the denominator.
Read the words after 'given' or the event after the bar first.
Without replacement, update the known removed colour and the total.
30-second recap
Choose, carry out, check
Translate the wording before you reach for a rule.
- Restrict first.
- Find the overlap within that group.
- Divide by the condition total.
- Check direction and replacement.
Quick answers
Conditional probability FAQ
Does 'given' always mean a previous event in time?
No. It can simply mean information about the same person or object, such as knowing they are a swimmer.
Do I need a tree diagram?
Only if it helps organise stages. Tables, Venn regions and lists can show the restricted group more directly.
Content standards
Curriculum and rights review
P9 conditional probability is Higher tier. Introductory counts support the definition before conditional notation and dependent draws. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.
Official specification references