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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).

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Student-friendly explanation

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.

How to write this in exams

  1. 1

    Start with the exact idea

    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. 2

    Then show how to use it

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

  3. 3

    Add one concrete 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.

  4. 4

    Avoid this incomplete answer

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

Definition

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).

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.

Rule to remember

Bayes' formula: P(Ei|A) = [P(Ei)P(A|Ei)]/[P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En)]. Conditions: E1, ..., En must be mutually exclusive and exhaustive, P(Ei)>0, and P(A)>0.

Memory hook

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

Examples and method

Worked example

A factory has machines M1 and M2 producing 70% and 30% of items. Defective rates are 3% and 8%. An item is found defective. Find the probability it was produced by M2. Let A be defective. P(M2|A)=P(M2)P(A|M2)/[P(M1)P(A|M1)+P(M2)P(A|M2)]. Substitute: 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. Therefore the probability is 8/15.

Method to apply

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

Diagram support

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.

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.

Avoid common mistakes

Common confusion

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

Common wrong answer

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

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.

Quick check

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.

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.

Answer writing and exam use

1-mark answer

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-mark answer

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). Bayes' formula: P(Ei|A) = [P(Ei)P(A|Ei)]/[P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En)]. Conditions: E1, ..., En must be mutually exclusive and exhaustive, P(Ei)>0, and P(A)>0. 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-mark answer

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. Bayes' formula: P(Ei|A) = [P(Ei)P(A|Ei)]/[P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En)]. Conditions: E1, ..., En must be mutually exclusive and exhaustive, P(Ei)>0, and P(A)>0. A factory has machines M1 and M2 producing 70% and 30% of items. Defective rates are 3% and 8%. An item is found defective. Find the probability it was produced by M2. Let A be defective. P(M2|A)=P(M2)P(A|M2)/[P(M1)P(A|M1)+P(M2)P(A|M2)]. Substitute: 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. Therefore the probability is 8/15. 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. Using P(A|M2)=0.08 as P(M2|A), which ignores the prior production share and the total defective probability.
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