Bayes’ theorem: partition, total probability and examples

What is Bayes’ theorem?

In plain words: if a ball is drawn from one of two bags, it is easy to find the chance of a red ball from a given bag. Bayes’ theorem answers the reverse question: given that the ball is red, how likely is it that it came from a particular bag (NCERT Section 13.5, page 423)? It rests on two ideas, a partition and the theorem of total probability.

Partition of a sample space

In other words, the events E1, E2, ..., En represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.

NCERT Mathematics Part II (Class 12), Section 13.5.1, page 424

In Class 11 words, the events are mutually exclusive and exhaustive. Any event E with 0 < P(E) < 1 and its complement E′ form a partition, and a partition is not unique (page 424).

NCERT slip (page 424): the book says that “any nonempty event E and its complement E′ form a partition”. If E is the whole sample space, E′ is empty and has probability 0, which a partition does not allow. In an exam, check that every part has a nonzero probability.

Theorem of total probability

If E_₁, E_₂, …, E_ₙ form a partition of S and A is any event, A splits into the disjoint pieces A ∩ E_ᵢ, and the multiplication rule gives each piece (NCERT Section 13.5.2, pages 424–425):

NCERT Example 15 (page 425): a strike has probability 0.65; a job is finished on time with probability 0.32 if there is a strike and 0.80 if not. So P(on time) = 0.65 × 0.32 + 0.35 × 0.80 = 0.488.

Bayes’ theorem

If E_₁, …, E_ₙ partition S and P(A) ≠ 0, then (NCERT page 425):

The proof (NCERT page 425) uses the multiplication rule on top and total probability below:

NCERT’s terms (Remark, page 426): the E_ᵢ are hypotheses, P(E_ᵢ) is the priori (a priori) probability of E_ᵢ, and P(E_ᵢ|A) is its a posteriori probability. The theorem is also called the formula for the probability of “causes”. Here a hypothesis need not be a physical cause: Bayes’ theorem updates probabilities from evidence and does not by itself establish causation.

On a tree, the bottom of the fraction is the sum of all the paths that end in A, and the top is the one path through E_ᵢ.

NCERT Example 18 (pages 427–428) shows why the answer can surprise: a test detects 90% of cases and gives a false positive for 1% of healthy people, and 0.1% of people have the disease. Then P(disease | positive) = (0.001 × 0.9)/(0.001 × 0.9 + 0.999 × 0.01) ≈ 0.083, because healthy people are so many more.

NCERT slip (page 423): the text says the problem was solved by “John Bayes”. Beyond the NCERT text (Wikipedia, “Thomas Bayes”, facts in our words): the theorem is named after Thomas Bayes (about 1701–1761), an English Presbyterian minister and statistician. He never published it; after his death his friend Richard Price brought the essay before the Royal Society, which heard it in 1763 and printed it in its Philosophical Transactions. The name does not affect the mathematics.

A random variable

NCERT ends the section with a remark (pages 430–431): a random variable is a real valued function whose domain is the sample space of a random experiment. For two coin tosses, X = the number of heads gives X(HH) = 2, X(HT) = X(TH) = 1 and X(TT) = 0, and Y = heads minus tails gives Y(HH) = 2, Y(HT) = Y(TH) = 0 and Y(TT) = −2: two random variables on the same sample space.

NCERT slip (page 406): the chapter’s Introduction promises a random variable, its probability distribution, mean and variance, and the Binomial distribution. The 2026–27 chapter stops here: Section 13.5 ends with this remark (pages 430–431). In an exam, follow Sections 13.2–13.5. Beyond the NCERT text: the JEE Main syllabus also lists the probability distribution of a random variable (NTA, JEE Main syllabus, Unit 13).

Solved examples

Common mistakes to avoid

Quick revision summary

• A partition: pairwise disjoint, exhaustive events with nonzero probabilities.
• Total probability: P(A) = P(E_₁) P(A|E_₁) + … + P(E_ₙ) P(A|E_ₙ), the sum of the paths to A.
• Bayes: P(E_ᵢ|A) = P(E_ᵢ) P(A|E_ᵢ) / [P(E_₁) P(A|E_₁) + … + P(E_ₙ) P(A|E_ₙ)], one path over all paths to A.
• A random variable is a real valued function on the sample space.

Questions students ask

What is the difference between conditional probability and Bayes’ theorem?

Conditional probability is the general idea, the chance P(E|A) of an event E once A is known to have occurred, while Bayes’ theorem is a formula for one kind of conditional probability: the reverse one, P(E_ᵢ|A), worked out from the forward probabilities P(A|E_ᵢ) and the prior probabilities P(E_ᵢ) of a partition. Every Bayes answer is a conditional probability, but not every conditional probability needs Bayes’ theorem: the chance of a king given a red card comes straight from counting, while the chance that a red ball came from a particular bag needs it, because what is known is the chance of red from each bag. NCERT calls this the “reverse probability” (page 423).

How do you know whether a question needs total probability or Bayes’ theorem?

A question needs the theorem of total probability when it asks for the overall chance of an outcome A that can come about in several ways, and Bayes’ theorem when it says that A has happened and asks for the chance of one of the ways, a “cause” E_ᵢ. Total probability adds all the paths to A on a tree, and Bayes’ theorem divides one of those paths by that total, so a Bayes answer needs the total first. If plant X makes 90% of a factory’s chips, 2% of them faulty, and plant Y makes 10%, 9% of them faulty, the faulty paths are 0.9 × 0.02 = 0.018 and 0.1 × 0.09 = 0.009, so P(faulty) is 0.027, and a faulty chip came from Y with probability 0.009/0.027 = 1/3.

Which chapter of NCERT Class 12 Maths has Bayes’ theorem?

Bayes’ theorem is in Chapter 13, Probability, of NCERT Mathematics Part II (Class 12), Section 13.5 (pages 423–433), which also covers the partition of a sample space and the theorem of total probability; the theorem itself is stated and proved on page 425. Its worked examples are NCERT Examples 16–21 (pages 426–430), followed by Exercise 13.3 (pages 431–433), and Miscellaneous Examples 22 and 24 use it again. The theorem is named after Thomas Bayes (beyond the NCERT text, whose page 423 has a slip in his first name), so it is written Bayes’ theorem (or Bayes’s theorem).

What are sensitivity and specificity in Bayes’ theorem questions on medical tests?

Beyond the NCERT text, medical statistics calls the chance that a test is positive for someone ill its sensitivity, and the chance that it is negative for a healthy person its specificity, so the false-positive rate is 1 minus the specificity; in NCERT Exercise 13.3, Q5, “99% effective” is a sensitivity. Bayes’ theorem turns them into the chance of disease given a positive result, the “predictive value positive” of the US CDC’s surveillance guidelines, which also depends on how common the disease is. With sensitivity 95%, specificity 90% and 2% of people ill, the positive paths are 0.019 and 0.098, so a positive result means disease with probability 0.019/0.117 = 19/117, about 0.16.

Do the probabilities given by Bayes’ theorem always add up to 1?

Yes, the a posteriori (posterior) probabilities P(E_₁|A), …, P(E_ₙ|A) of a partition always add up to 1 when P(A) ≠ 0, because they share the denominator P(A), and their numerators P(E_ᵢ) P(A|E_ᵢ) add up to P(A) by the theorem of total probability: once A has happened, exactly one of the partition events has occurred. This gives a check on the working, and the last posterior can be found by subtracting the others from 1. The priors P(E_ᵢ) add up to 1 as well, but the forward probabilities P(A|E_ᵢ) need not: in NCERT Example 17 (page 427) the chances of a gold coin from the three boxes are 1, 0 and 1/2.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. An urn has 5 red and 5 black balls. A ball is drawn, its colour noted, and it is returned with 2 more balls of that colour. Find the probability that a second draw gives red (NCERT Exercise 13.3, Q1, page 431).

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   By total probability: (1/2)(7/12) + (1/2)(5/12) = 1/2.

2. Q2. An insurer covers 2000 scooter drivers, 4000 car drivers and 6000 truck drivers, with accident probabilities 0.01, 0.03 and 0.15. An insured person has an accident. Find the probability that he drives a scooter (NCERT Exercise 13.3, Q7, page 432).

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   Expected accidents: 20, 120 and 900, so the probability is 20/1040 = 1/52.

3. Q3. Machine A makes 60% of the items and B 40%; 2% of A’s and 1% of B’s items are defective. A defective item is found. Find the probability that B made it (NCERT Exercise 13.3, Q8, page 432).

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   0.4 × 0.01 / (0.6 × 0.02 + 0.4 × 0.01) = 0.004/0.016 = 1/4.

4. Q4. If A ⊂ B and P(B) ≠ 0, which is correct? (A) P(A|B) = P(B)/P(A); (B) P(A|B) < P(A); (C) P(A|B) ≥ P(A); (D) none of these (NCERT Exercise 13.3, Q14, page 433).

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   P(A|B) = P(A ∩ B)/P(B) = P(A)/P(B) ≥ P(A), since P(B) ≤ 1: (C).