Conditional Probability from Tables, Tree Diagrams and Venn Diagrams Revision Notes | GCSE AQA Higher Maths
Conditional probability, written P(A | B), is the probability of event A occurring given that event B has already occurred. Free GCSE Maths notes.
Updated 24 Mar 2026 · Revise Right Team
Conditional Probability from Tables, Tree Diagrams and Venn Diagrams is part of Conditional Probability in GCSE AQA Higher Maths.
3 min
Reading time
12
Lessons in this topic
GCSE AQA Higher
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
Conditional probability, written P(A | B), is the probability of event A occurring given that event B has already occurred. Free GCSE Maths notes.
Explain how conditional probability changes the sample space when extra information is given
Calculate conditional probability from a two-way table by choosing the correct restricted total
Calculate conditional probability from a tree diagram for independent and dependent events
Calculate conditional probability from a Venn diagram and interpret the result in context
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Use this lesson to stay inside Conditional Probability while connecting the detail back to the wider Probability topic.
Conditional probability: The probability of an event occurring given that another event is already known to have occurred.
Restricted sample space: The reduced set of possible outcomes that remains once a condition has been applied.
Two-way table: A table that organises data by two categories and helps identify the correct denominator for conditional probability.
Dependent event: An event whose probability changes because a previous event has already happened.
Revision notes
Read the core explanations from Conditional Probability from Tables, Tree Diagrams and Venn Diagrams before testing yourself from memory.
1. Why Conditional Probability Changes Everything
Here is a question that trips up a lot of students: a bag contains 3 red and 2 blue counters. You pick one and keep it. What is the probability the second counter is red? The answer depends entirely on what the first counter was — and that dependence is exactly what conditional probability captures.
Conditional probability is not a new formula to memorise. It is a shift in perspective: instead of asking 'what is the probability of this event?', you ask 'what is the probability of this event, now that I know something else?' That extra information shrinks the sample space, and the probability changes accordingly.
At GCSE, you will not need to use the formal conditional probability formula. Instead, the examiner expects you to read conditional probabilities from three representations: two-way tables, tree diagrams, and Venn diagrams. Each one makes the restricted sample space visible in a different way.
2. Two-Way Tables and Conditional Probability
A two-way table is the most direct tool for conditional probability. The trick is knowing which total to use as your denominator. When a condition is given, your denominator is the row total or column total for that condition — not the grand total.
The table above shows results for 50 students. To find $P(\text{Passed} \mid \text{Studied})$, you restrict attention to the 'Studied' row only. There are $28 + 7 = 35$ students who studied, and 28 of them passed.
Worked example: Reading Conditional Probability from a Two-Way Table
Using the table above, calculate $P(\text{Passed} \mid \text{Studied})$ and $P(\text{Studied} \mid \text{Passed})$.
Condition: 'Studied' — restrict to the Studied row. — The condition tells you which row or column becomes your new sample space. [M1]
$28 + 7 = 35$ students studied in total. — This is the denominator for any probability conditional on 'Studied'.
28 students studied AND passed. — This is the numerator — the count satisfying both the condition and the event.
Condition is now 'Passed' — restrict to the Passed column. Column total: $28 + 6 = 34$. Favourable: 28 studied and passed. $P(\text{Studied} \mid \text{Passed}) = \dfrac{28}{34} = \dfrac{14}{17}$ — Notice the denominator changed — the condition changed, so the restricted sample space changed. [A1]
Notice that $P(\text{Passed} \mid \text{Studied})
eq P(\text{Studied} \mid \text{Passed})$. Swapping the condition and the event gives a completely different probability. This is one of the most common errors on GCSE Higher exam papers.
3. Tree Diagrams for Dependent Events
When events happen in sequence and the outcome of the first affects the second, a tree diagram makes the conditional probabilities explicit. Each second-stage branch already has the condition built in — the probabilities on those branches are conditional probabilities.
The key rule for tree diagrams: multiply along branches to find the probability of a combined outcome, and add the results for different routes to the same overall outcome.
In the tree above, the bag starts with 3 red and 2 blue counters (5 total). After a red is removed, only 4 counters remain — 2 red and 2 blue. After a blue is removed, 4 remain — 3 red and 1 blue. The second-branch probabilities change depending on what happened first: that is conditional probability in action.
Worked example: Using a Tree Diagram to Find a Conditional Probability
A bag has 3 red and 2 blue counters. Two counters are picked without replacement. Given that the second counter is red, find the probability that the first counter was also red.
Route 1: Red then Red (RR)
Route 2: Blue then Red (BR) — These are the only two routes that end with a red second counter. The condition restricts us to these routes only. [M1]
$P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ — First pick: 3 red out of 5. Second pick: 2 red remain out of 4 total remaining.
$P(BR) = \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20}$ — First pick: 2 blue out of 5. Second pick: 3 red remain out of 4 total remaining.
$P(\text{2nd is red}) = \dfrac{6}{20} + \dfrac{6}{20} = \dfrac{12}{20}$ — Add the two routes — this is the denominator for the conditional probability. [M1]
$P(\text{1st red} \mid \text{2nd red}) = \dfrac{P(RR)}{P(\text{2nd red})} = \dfrac{6/20}{12/20} = \dfrac{6}{12} = \dfrac{1}{2}$ — Divide the favourable route by the total probability of the condition. [A1]
A Venn diagram shows how two sets overlap. For conditional probability, the logic is the same as with a two-way table: the condition tells you which region becomes your new total, and you count how much of the event falls inside that region.
Worked example: Finding Conditional Probability from a Venn Diagram
Using the Venn diagram above (40 students total), calculate $P(T \mid F)$ and interpret your answer.
Condition: plays football ($F$). Restrict to the entire football circle. — The football circle contains all students for whom the condition is true. [M1]
$15 + 8 = 23$ students play football. — This is the denominator — the size of the restricted sample space.
8 students play both football and tennis (the intersection). — These are the students who satisfy both the condition (plays football) and the event (plays tennis).
$P(T \mid F) = \dfrac{8}{23}$ — Numerator: intersection. Denominator: total in the condition's circle. [A1]
$\dfrac{8}{23} \approx 0.348$ — Among students who play football, approximately 34.8% also play tennis. Always relate the answer back to the context when asked to interpret. [A1]
Answer: $P(T \mid F) = \dfrac{8}{23}$. Among students who play football, the probability they also play tennis is $\dfrac{8}{23}$.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-24.
Conditional probability, written P(A | B), is the probability of event A occurring given that event B has already occurred. The condition restricts the sample space to outcomes where B has occurred, so the probability changes.
How do you find conditional probability from a Venn diagram?
Divide the intersection by the total in the circle for the condition. If 23 students play football and 8 play both football and tennis, P(T | F) = 8/23 ≈ 0.348.
How do you find conditional probability from a tree diagram?
Add the routes that satisfy the condition to get the denominator, then divide the favourable route by it. With 3 red and 2 blue counters picked without replacement, P(RR) = 6/20 and P(BR) = 6/20, so P(1st red | 2nd red) = 6/12 = 1/2.
How do you find conditional probability from a two-way table?
Use the row or column total for the condition as the denominator, not the grand total. If 35 students studied and 28 of them passed, P(Passed | Studied) = 28/35 = 4/5.
Is P(A | B) the same as P(B | A)?
P(A | B) and P(B | A) are not the same, because swapping the condition and the event changes the denominator. P(Passed | Studied) = 28/35 = 4/5, but P(Studied | Passed) = 28/34 = 14/17.
How does conditional probability change the sample space?
Conditional probability restricts the sample space to only the outcomes where the given condition has happened. The extra information shrinks the group being looked at, so the probability changes accordingly.
Which total do you use for conditional probability from a table?
For conditional probability from a two-way table, the denominator is the row or column total for the condition, not the grand total. For P(Passed | Studied), the denominator is the 35 students who studied.
What does the vertical line mean in probability?
The vertical line in P(A | B) is read as 'given', so it means the probability of A given that B has occurred. The event after the line is the condition that restricts the sample space.
Given the second counter is red, what is the chance the first was red?
With 3 red and 2 blue counters picked without replacement, P(RR) = 3/5 × 2/4 = 6/20 and P(BR) = 2/5 × 3/4 = 6/20. So P(second red) = 12/20, and P(first red | second red) = 6/12 = 1/2.
Why are second-branch probabilities on a tree diagram conditional?
Second-branch probabilities on a tree diagram are conditional because each one already has the first outcome built in. After a red is removed from 3 red and 2 blue, the branches show 2 red and 2 blue out of 4.