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Theorem of Total Probability

If E1, E2, ..., En are mutually exclusive and exhaustive events with P(Ei)>0, then for any event A, P(A)=Σ P(Ei)P(A|Ei).

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Student-friendly explanation

The events E1 to En divide the sample space into non-overlapping cases. To find P(A), find the probability of A through each case and add all such contributions. This theorem is used when A can happen through several possible sources or groups.

How to write this in exams

  1. 1

    Start with the exact idea

    If E1, E2, ..., En are mutually exclusive and exhaustive events with P(Ei)>0, then for any event A, P(A)=Σ P(Ei)P(A|Ei).

  2. 2

    Then show how to use it

    List all possible cases E1, E2, ..., En. Verify they are mutually exclusive and exhaustive. Write P(Ei) for each case. Write P(A|Ei) for each case. Multiply case probability by conditional probability and add all terms.

  3. 3

    Add one concrete example

    A product comes from machines M1 and M2. If P(M1)=0.6, P(M2)=0.4, P(defective|M1)=0.02, and P(defective|M2)=0.05, then P(defective)=0.6×0.02+0.4×0.05=0.032.

  4. 4

    Avoid this incomplete answer

    Using only the largest case or ignoring one case, which makes the total probability incomplete.

Definition

If E1, E2, ..., En are mutually exclusive and exhaustive events with P(Ei)>0, then for any event A, P(A)=Σ P(Ei)P(A|Ei).

Example

A product comes from machines M1 and M2. If P(M1)=0.6, P(M2)=0.4, P(defective|M1)=0.02, and P(defective|M2)=0.05, then P(defective)=0.6×0.02+0.4×0.05=0.032.

Rule to remember

Key theorem: If E1, E2, ..., En form a partition of S, then P(A)=P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En). Conditions: Ei∩Ej=∅ for i≠j, union of all Ei is S, and P(Ei)>0 for conditioning.

Memory hook

Break into cases, multiply within each case, add across cases.

Examples and method

Worked example

A bag is chosen at random. Bag I is chosen with probability 1/3 and contains a red ball with probability 2/5. Bag II is chosen with probability 2/3 and contains a red ball with probability 3/4. Find probability of drawing a red ball. Let A be red. P(A)=P(I)P(A|I)+P(II)P(A|II)=(1/3)(2/5)+(2/3)(3/4)=2/15+1/2=4/30+15/30=19/30.

Method to apply

List all possible cases E1, E2, ..., En. Verify they are mutually exclusive and exhaustive. Write P(Ei) for each case. Write P(A|Ei) for each case. Multiply case probability by conditional probability and add all terms.

Diagram support

A partition diagram or probability tree is useful. The sample space may be divided into cases E1, E2, ..., En, each leading to A with a conditional probability.

How CBSE asks it

It appears before Bayes' theorem, often as the denominator in reverse probability questions. It is also asked directly in manufacturing, medical testing, bag selection, and population group problems.

Avoid common mistakes

Common confusion

Students may add conditional probabilities directly, such as P(A|E1)+P(A|E2), without multiplying by the probabilities of the cases.

Common wrong answer

Using only the largest case or ignoring one case, which makes the total probability incomplete.

Exam tip

First check that the cases form a partition: no overlap and together cover the whole sample space.

Quick check

A test sample is from factory F1 with probability 0.7 and F2 with probability 0.3. If P(reject|F1)=0.04 and P(reject|F2)=0.08, find P(reject).

P(reject)=0.7×0.04+0.3×0.08=0.028+0.024=0.052.

Answer writing and exam use

1-mark answer

If E1, E2, ..., En are mutually exclusive and exhaustive events with P(Ei)>0, then for any event A, P(A)=Σ P(Ei)P(A|Ei).

2-mark answer

If E1, E2, ..., En are mutually exclusive and exhaustive events with P(Ei)>0, then for any event A, P(A)=Σ P(Ei)P(A|Ei). Key theorem: If E1, E2, ..., En form a partition of S, then P(A)=P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En). Conditions: Ei∩Ej=∅ for i≠j, union of all Ei is S, and P(Ei)>0 for conditioning. A product comes from machines M1 and M2. If P(M1)=0.6, P(M2)=0.4, P(defective|M1)=0.02, and P(defective|M2)=0.05, then P(defective)=0.6×0.02+0.4×0.05=0.032.

3-mark answer

The events E1 to En divide the sample space into non-overlapping cases. To find P(A), find the probability of A through each case and add all such contributions. This theorem is used when A can happen through several possible sources or groups. Key theorem: If E1, E2, ..., En form a partition of S, then P(A)=P(E1)P(A|E1)+P(E2)P(A|E2)+...+P(En)P(A|En). Conditions: Ei∩Ej=∅ for i≠j, union of all Ei is S, and P(Ei)>0 for conditioning. A bag is chosen at random. Bag I is chosen with probability 1/3 and contains a red ball with probability 2/5. Bag II is chosen with probability 2/3 and contains a red ball with probability 3/4. Find probability of drawing a red ball. Let A be red. P(A)=P(I)P(A|I)+P(II)P(A|II)=(1/3)(2/5)+(2/3)(3/4)=2/15+1/2=4/30+15/30=19/30. It appears before Bayes' theorem, often as the denominator in reverse probability questions. It is also asked directly in manufacturing, medical testing, bag selection, and population group problems. Using only the largest case or ignoring one case, which makes the total probability incomplete.
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