medium difficulty8 min

Bayes' Theorem: Reverse Conditional Probability

If E1, E2, ..., En form a partition of the sample space and A is an event with P(A)>0, then P(Ei|A)=P(Ei)P(A|Ei)/ΣP(Ej)P(A|Ej).

In one line: Bayes asks: after seeing the result, which source is most likely?

Practice This Concept

Understand it

Bayes' theorem finds the probability of a cause or case after an observed event has occurred. The numerator is the contribution of the required case to event A. The denominator is the total probability of A from all cases.

Example

If 60% items are from machine M1 and 40% from M2, with defective rates 2% and 5%, then the probability that a defective item came from M2 is (0.4×0.05)/(0.6×0.02+0.4×0.05)=0.02/0.032=5/8.

Formula and rule

Conditions: E1, ..., En must be mutually exclusive and exhaustive, P(Ei)>0, and P(A)>0.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A factory has machines M1 and M2 producing 70% and 30% of items.

1Defective rates are 3% and 8%.
2An item is found defective.
3Find the probability it was produced by M2.
4Let A be defective.
5P(M2|A)=P(M2)P(A|M2)/[P(M1)P(A|M1)+P(M2)P(A|M2)].
6Substitute: P(M2|A)=(0.30)(0.08)/[(0.70)(0.03)+(0.30)(0.08)] = 0.024/(0.021+0.024)=0.024/0.045=8/15.
7Therefore the probability is 8/15.

Method you can reuse

1Identify the observed event A.
2List all possible causes or cases E1 to En.
3Check that the cases form a partition.
4Compute each path probability P(Ei)P(A|Ei).
5Add all path probabilities to get P(A).
6Put the required path probability in the numerator and divide by P(A).

Visual tip

A probability tree is strongly useful. First-level branches show the possible sources E1, E2, ..., En; second-level branches show A and not A under each source. The A-branches combine to form the denominator.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Two boxes are chosen with probabilities 1/4 and 3/4. The probability of a red ball from them is 1/2 and 1/3 respectively. If a red ball is drawn, find the probability it came from the first box.

Think first, then show answer

P(Box1|Red)=((1/4)(1/2))/((1/4)(1/2)+(3/4)(1/3))=(1/8)/(1/8+1/4)=(1/8)/(3/8)=1/3.

Common wrong answer

Using P(A|M2)=0.08 as P(M2|A), which ignores the prior production share and the total defective probability.

Don't mix this up

Students often put P(A|Ei) as the final answer instead of finding P(Ei|A). Bayes' theorem reverses the condition.

Exam tip

Name the observed event as A and the possible sources as E1, E2, ..., En. The required source after observation goes in the numerator.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

It is asked in diagnostic tests, machine defects, bag selection, population groups, and case-study data. The required probability is usually a reverse condition after an observed result.

1
Enough for 1 mark
If E1, E2, ..., En form a partition of the sample space and A is an event with P(A)>0, then P(Ei|A)=P(Ei)P(A|Ei)/ΣP(Ej)P(A|Ej).
2
Add for 2 marks
Example: If 60% items are from machine M1 and 40% from M2, with defective rates 2% and 5%, then the probability that a defective item came from M2 is (0.4×0.05)/(0.6×0.02+0.4×0.05)=0.02/0.032=5/8.
3
Add for 3 marks
Bayes' theorem finds the probability of a cause or case after an observed event has occurred.

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