Conditional Probability and Venn Diagrams
Conditional Probability and Venn Diagrams
Conditional probability deals with the probability of an event occurring given that another event has already occurred. Combined with Venn diagrams and tree diagrams, it provides powerful tools for solving complex probability problems.
Conditional Probability Formula
The probability of A given B (the probability of A occurring, knowing that B has occurred) is:
P(A|B) = P(A ∩ B) / P(B)
where P(A ∩ B) is the probability that both A and B occur, and P(B) > 0.
Rearranging gives the multiplication rule:
P(A ∩ B) = P(A|B) × P(B) = P(B|A) × P(A)
Worked Example: In a class of 30 students, 18 study Maths, 12 study Physics, and 8 study both. A student is chosen at random. Given that they study Maths, find the probability they also study Physics.
P(Physics | Maths) = P(Physics ∩ Maths) / P(Maths) = (8/30) / (18/30) = 8/18 = 4/9 ≈ 0.444
Venn Diagrams
A Venn diagram represents events as overlapping circles within a rectangle (the sample space). For two events A and B:
- The overlap region represents A ∩ B (A and B both occur)
- A only (not B) is the part of circle A outside the overlap
- The region outside both circles represents (A ∪ B)' (neither A nor B)
Setting up a Venn diagram: Always fill in the intersection first, then work outward.
Worked Example: In a survey of 100 people, 60 like coffee, 45 like tea, and 20 like both. Illustrate with a Venn diagram and find the probability that a randomly chosen person likes neither.
| Region | Count |
|---|---|
| Coffee only | 60 - 20 = 40 |
| Tea only | 45 - 20 = 25 |
| Both | 20 |
| Neither | 100 - 40 - 25 - 20 = 15 |
P(neither) = 15/100 = 0.15
The Addition Rule
For any two events:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
For mutually exclusive events (A ∩ B = ∅):
P(A ∪ B) = P(A) + P(B)
Independent Events
Two events are independent if the occurrence of one does not affect the probability of the other:
P(A|B) = P(A) (equivalently, P(A ∩ B) = P(A) × P(B))
To test for independence, check whether P(A ∩ B) = P(A) × P(B). If they are equal, the events are independent. If not, they are dependent.
Worked Example: A card is drawn from a standard deck. Let A = "card is red" and B = "card is a king".
P(A) = 26/52 = 1/2, P(B) = 4/52 = 1/13, P(A ∩ B) = 2/52 = 1/26.
P(A) × P(B) = (1/2)(1/13) = 1/26 = P(A ∩ B). So A and B are independent.
Tree Diagrams
Tree diagrams display sequential events with conditional probabilities on the branches. Multiply along branches for joint probabilities; add across final outcomes for combined probabilities.
Worked Example: Box A contains 3 red and 2 blue balls. Box B contains 4 red and 6 blue balls. A box is chosen at random, then a ball is drawn. Find the probability of drawing a red ball.
First branch: P(Box A) = 1/2, P(Box B) = 1/2.
Second branches: P(Red | A) = 3/5, P(Red | B) = 4/10 = 2/5.
P(Red) = P(Red ∩ A) + P(Red ∩ B)
= P(Red|A)×P(A) + P(Red|B)×P(B)
= (3/5)(1/2) + (2/5)(1/2)
= 3/10 + 2/10 = 1/2
Bayes' Theorem (the Reverse Conditional)
Given the result, what is the probability of a particular cause? This reverses the direction of a tree diagram:
P(A|B) = P(B|A) × P(A) / P(B)
This is often applied using a tree diagram — compute the probability of the observed outcome via each route, then find the proportion via the route of interest.
Worked Example (continuing above): Given that the ball drawn is red, find the probability it came from Box A.
P(A | Red) = P(Red ∩ A) / P(Red) = (3/10) / (1/2) = (3/10) × (2/1) = 3/5
Three-Event Venn Diagrams
For three events A, B, C, the Venn diagram has seven regions plus the exterior. Fill in from the centre outward:
1. Start with A ∩ B ∩ C
2. Then the pairwise intersections minus the triple
3. Then each circle minus all intersections
4. Finally the exterior
Exam Tips
- In Venn diagram problems, always calculate the intersection first and work outward.
- When asked "are events independent?", you must calculate and compare P(A ∩ B) with P(A)P(B). Do not just state it.
- Tree diagrams are essential for sequential events with conditional probabilities. Label every branch with its probability.
- For reverse conditional questions ("given the result, find the cause"), use the tree diagram and the formula P(cause|result) = P(result via cause) / P(result overall).
- Mutually exclusive events are not independent (unless one has probability zero). Students commonly confuse these concepts.
- P(A') = 1 - P(A). Use this complement rule to simplify "at least one" problems: P(at least one) = 1 - P(none).