Conditional Probability and Venn Diagrams
Conditional probability tells you how likely one event is once you already know that a related event has happened, and a two-set Venn diagram is the clearest way to see it. On this page you will learn to read probabilities straight from the regions of a Venn diagram, shrink the universe to work out P(A given B), and answer set-notation questions in the style of the Edexcel IGCSE Maths exam. When you are ready, scroll down to the auto-marked practice questions, where every card comes with a freshly generated diagram.
How to Find Probability from a Venn Diagram
A two-set Venn diagram splits everyone into exactly four regions. If every outcome is equally likely, probability becomes a counting exercise, and each step below has a picture of what it is asking you to look at.
Step 1. Know the four regions
Four regions, and every person is in exactly one of them.
In A only, in both A and B (the overlap), in B only, and in neither circle. Every person sits in exactly one of them, so nobody is counted twice and nobody is missed.
Step 2. Total first, and do not skip the outside
Every region counted, including the 7 outside: \(n(\xi) = 32\).
Add all four regions to get \(n(\xi)\), the number in the universal set. The region outside the circles is the one people forget, and it changes every denominator on the page.
Step 3. Spot the region the event describes
Shaded: in A or B. That is three regions, \(12 + 5 + 8 = 25\).
Decide exactly which region, or combination of regions, the question is asking about, then add those counts. Here "in A or B" is three regions together, not two.
Step 4. A whole circle includes the overlap
Shaded: the whole of circle A, \(n(A) = 12 + 5 = 17\). Circle B is not shaded.
\(n(A)\) is the A-only region plus the middle. That "plus the overlap" is where most marks are lost. Then divide: probability is the favourable count over the total count.
Core ideas
- \(A \cap B\) means "and": the overlap.
- \(A \cup B\) means "or": either circle, counted once.
- \(A'\) means "not in A".
Conditional Probability: Shrink the Universe
A question that says "given that..." is a conditional probability question. The extra information means you are no longer choosing from the whole diagram; you are choosing from a subset of it. That subset becomes your new universal set:
Given B, only circle B is left: 5 of the \(5 + 8 = 13\) in it are also in A.
- Find the circle (or region) for the event you are told has happened. Its total is the new denominator.
- Inside that circle, count how many are also in the event you are asked about.
- Divide the second count by the first.
The vertical bar reads "given": \(P(A \mid B)\) is the probability of A, given that B has happened. Notice the denominator is \(n(B)\), not \(n(\xi)\). The third worked example below shows this as a picture: the shaded circle is the shrunken universe.
Set Notation on a Venn Diagram
Three symbols do most of the work in this topic. Each one is a shaded picture before it is a calculation, so read the shape first and count second.
\(A \cap B\) means "A and B": only the overlap, the people in both circles.
\(A \cup B\) means "A or B": everything in either circle, with the overlap counted once, not twice.
\(A'\) means "not in A": everything outside circle A, including the region outside both circles.
Worked Examples
All four examples use the same survey: 40 students were asked whether they take photography (P) or textiles (T). 14 take photography only, 6 take both, 11 take textiles only and 9 take neither. Watch the same diagram evolve as the questions change.
💡 Example 1: Read the diagram
Find (a) the number of students who take photography, (b) the probability that a student picked at random takes photography.
The four regions and their counts
What's happening?
- Circle P is two regions: photography only (14) plus the overlap (6).
- The total is all four regions: 14, 6, 11 and 9 make 40.
- Probability is the circle count over the total count.
💡 Example 2: "And" and "or"
Find (a) \(P(P \cap T)\), (b) \(P(P \cup T)\).
The overlap \(P \cap T\) shaded
What's happening?
- \(P \cap T\) is the shaded overlap only: 6 students.
- \(P \cup T\) is everything inside the circles, each region counted once.
- Adding \(P(P)\) and \(P(T)\) would count the overlap twice; read the union from the diagram instead.
💡 Example 3: Conditional probability, the key picture
Given that a student takes photography, find the probability that they also take textiles. In symbols: work out \(P(T \mid P)\).
Circle P becomes the new universe; everything outside it is greyed out
What's happening?
- "Given that a student takes photography" means the student is somewhere inside circle P, so the 11 and the 9 are out of the question.
- The universe shrinks from 40 students to the 20 in circle P.
- Of those 20, the 6 in the overlap also take textiles, so the answer is 6 out of 20.
- Compare: \(P(T) = \dfrac{17}{40}\), but \(P(T \mid P) = \dfrac{6}{20}\). The condition changes the answer.
💡 Example 4: Complements
Find (a) \(P(P')\), (b) \(P(P' \cap T')\).
The complement \(P'\): everything outside circle P shaded
\(P' \cap T'\) is the region outside both circles, so \(P(P' \cap T') = \dfrac{9}{40}\).
What's happening?
- \(P'\) is every region outside circle P: textiles only (11) and neither (9).
- \(P' \cap T'\) means "not in P and not in T": only the neither region qualifies.
- Check: \(P(P)\) and \(P(P')\) add to 1.
Key Points and Common Pitfalls
🔑 Key points
- The four regions of a two-set Venn diagram always add up to \(n(\xi)\).
- A whole circle is its "only" region plus the overlap, so \(n(B)\) is the B-only count added to \(n(A \cap B)\).
- For \(P(A \mid B)\), divide the overlap by \(n(B)\): the circle you are given becomes the new universe.
- \(P(A \mid B)\) and \(P(B \mid A)\) usually have different denominators, so check which event is known.
- An unsimplified fraction is still a correct probability; simplify only if the question asks.
⚠️ Pitfalls
- Forgetting the "neither" region when adding up the total.
- Adding \(P(A)\) and \(P(B)\) to get \(P(A \cup B)\): that counts the overlap twice. Read the union off the diagram.
- Dividing by \(n(\xi)\) in a "given that" question. The denominator is the count of the known event, such as \(n(B)\).
- Treating \(A'\) as if it were just circle B: \(A'\) also includes the region outside both circles.