What is Bayes’ theorem in simple terms?

It finds the probability of a cause given that an effect has been observed. Multiply along each branch of the tree that leads to the effect, then divide the branch you want by the total of all such branches.

Do I have to draw a tree diagram?

It is not compulsory, but the curriculum notes recommend it for problems with a finite number of outcomes, and it makes every probability you need visible in one place.

If A and B are independent, are A′ and B independent too?

Yes. If A and B are independent then so are A′ and B, A and B′, and A′ and B′. M1 expects you to recognise this.

Conditional Probability and Bayes’ Theorem (DSE M1)

The conditional probability of A given B is P(A | B) = P(A ∩ B) / P(B). Bayes’ theorem reverses the condition: P(B | D) = P(B)P(D | B) / P(D), where P(D) is the total over every branch of a tree diagram that ends in D.

Solve Conditional Probability and Bayes’ Theorem questions with AI

GeoGebra tree diagram with branches for machine A 0.6 and machine B 0.4, then defective and not defective, used for Bayes theorem — The tree diagram for Example 1. The two red branches end in a defective item; machine B’s share of them is 0.020 / 0.032 = 0.625. Try it yourself →

Conditional Probability Formulas

Conditional probabilityP(A | B) = P(A ∩ B) / P(B)

Multiplication ruleP(A ∩ B) = P(A) P(B | A)

Total probabilityP(D) = P(A)P(D | A) + P(B)P(D | B)

Bayes’ theoremP(B | D) = P(B)P(D | B) / [P(A)P(D | A) + P(B)P(D | B)]

Independent eventsP(A ∩ B) = P(A)P(B) ⟺ P(A | B) = P(A)

Worked Examples — Conditional Probability and Bayes’ Theorem

Example 1: Bayes’ theorem with a tree diagram

A factory has two machines. Machine A makes 60% of the items and machine B makes 40%. 2% of the items from A and 5% of the items from B are defective. An item chosen at random is defective. Find the probability that it was made by machine B.

P(A ∩ D) = 0.6 × 0.02 = 0.012
P(B ∩ D) = 0.4 × 0.05 = 0.020
P(D) = 0.012 + 0.020 = 0.032
P(B | D) = 0.020 / 0.032 = 0.625

Answer: 0.625

Example 2: Testing whether two events are independent

Given P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7, determine whether A and B are independent.

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2
P(A) × P(B) = 0.5 × 0.4 = 0.2
The two are equal, so A and B are independent.

Answer: Independent, since P(A ∩ B) = P(A)P(B).

💡 Exam Traps & Secrets

⚠️ P(B | D) is not P(D | B)In Example 1, P(D | B) = 0.05 but P(B | D) = 0.625. Read which event is known to have happened: that one goes after the bar and in the denominator.

⚠️ Independent is not mutually exclusiveMutually exclusive events cannot happen together, so P(A ∩ B) = 0. Independent events can; one simply does not change the probability of the other.

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Bayes’ theorem)A test detects a disease with probability 0.95 and gives a false positive with probability 0.02. 1% of people have the disease. Find the probability that a person who tests positive has the disease.

(Tree diagram)Bag X has 3 red and 2 blue balls; bag Y has 1 red and 4 blue. A bag is chosen at random and a red ball is drawn. Find the probability it came from bag X.

(Independence)If A and B are independent with P(A) = 0.3 and P(B) = 0.6, find P(A ∪ B) and P(A′ ∩ B).