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CraftExam
medium difficulty7 min

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).

In one line: Break into cases, multiply within each case, add across cases.

Practice This Concept

Understand it

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.

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.

Formula and rule

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.

See it worked out

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

Worked example

A bag is chosen at random.

  1. 1Bag I is chosen with probability 1/3 and contains a red ball with probability 2/5.
  2. 2Bag II is chosen with probability 2/3 and contains a red ball with probability 3/4.
  3. 3Find probability of drawing a red ball.
  4. 4Let A be red.
  5. 5P(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 you can reuse

  1. 1List all possible cases E1, E2, ..., En.
  2. 2Verify they are mutually exclusive and exhaustive.
  3. 3Write P(Ei) for each case.
  4. 4Write P(A|Ei) for each case.
  5. 5Multiply case probability by conditional probability and add all terms.

Visual tip

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.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

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).

Think first, then show answer
P(reject)=0.7×0.04+0.3×0.08=0.028+0.024=0.052.

Common wrong answer

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

Don't mix this up

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

Exam tip

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

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 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.
  1. 1

    Enough for 1 mark

    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

    Add for 2 marks

    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.

  3. 3

    Add for 3 marks

    The events E1 to En divide the sample space into non-overlapping cases.

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