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Bayes questions read like puzzles: a ball moves between bags, a witness may be lying, a test may be wrong. The arithmetic is short once the set-up is right, and most errors are in the set-up. This post gives a three-step method, works three different problem types, and ends with a check that catches most wrong answers.
Where is this in the syllabus?
The NCERT Class 12 chapter on probability covers conditional probability in section 13.2 and the partition of a sample space, the theorem of total probability and Bayes' theorem in section 13.5. The CBSE Mathematics curriculum for 2026-27 lists "Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes' theorem". The JEE (Main) 2026 syllabus and the JEE Advanced 2026 syllabus include it too. Check the current year's syllabus.
What does conditional probability mean?
P(A | B) is the probability of A once you know B has happened: P(A | B) = P(A ∩ B) ÷ P(B). In practice, knowing B shrinks the sample space to B.
Example. Two dice are thrown and the sum is 8. What is the probability that at least one die shows 6? The outcomes with sum 8 are (2,6), (3,5), (4,4), (5,3) and (6,2): five outcomes. Two contain a 6. So the answer is 2/5.
What does Bayes' theorem add?
Often you know the probability of a result given each cause, and want the reverse: the probability of a cause given the result. NCERT calls the causes E₁, E₂, …, Eₙ "hypotheses", P(Eᵢ) the prior probability and P(Eᵢ | A) the a posteriori probability. The causes must form a partition: exactly one of them happens.
P(Eᵢ | A) = P(Eᵢ)P(A | Eᵢ) ÷ [P(E₁)P(A | E₁) + … + P(Eₙ)P(A | Eₙ)]
The denominator is P(A), from the theorem of total probability.
The three-step method
List the causes and their priors. Check that the priors add to 1.
Write the likelihood of the result under each cause, P(A | Eᵢ).
Multiply along each branch, add the branches, divide. The branch you want goes on top; the sum goes below.
Worked example 1: a transferred ball
Bag X has 3 red and 2 blue balls. Bag Y has 2 red and 3 blue. One ball is moved at random from X to Y, then one ball is drawn from Y. It is red. What is the probability that the moved ball was red?
Causes: moved red, P = 3/5; moved blue, P = 2/5. They add to 1.
If red was moved, Y has 3 red and 3 blue, so P(red) = 3/6 = 1/2. If blue was moved, Y has 2 red and 4 blue, so P(red) = 2/6 = 1/3.
Branches: 3/5 × 1/2 = 3/10 = 9/30, and 2/5 × 1/3 = 2/15 = 4/30. Total = 13/30. Answer: (9/30) ÷ (13/30) = 9/13.
Check: the other posterior is 4/13, and 9/13 + 4/13 = 1.
Worked example 2: a report that may be false
A person tells the truth 2 out of 3 times. They throw a die and report a six. What is the probability it really was a six? Assume that when they lie about a non-six, they report six.
Causes: six, P = 1/6; not six, P = 5/6.
P(report six | six) = 2/3 (truth). P(report six | not six) = 1/3 (lie).
Branches: 1/6 × 2/3 = 2/18, and 5/6 × 1/3 = 5/18. Answer: 2/18 ÷ 7/18 = 2/7.
Even a fairly honest reporter leaves only a 2/7 chance, because sixes are rare to begin with. The prior matters as much as the reliability.
Worked example 3: thinking in counts
20% of emails are spam. A filter flags 90% of spam and 5% of genuine emails. An email is flagged. What is the chance it is spam?
Imagine 1000 emails: 200 spam, of which 180 are flagged; 800 genuine, of which 40 are flagged. Of 220 flagged emails, 180 are spam: 180/220 = 9/11 ≈ 0.82. The formula gives the same: 0.18 ÷ (0.18 + 0.04) = 9/11.
Counting like this is a teachable skill. Sedlmeier and Gigerenzer (2001) compared training people to build frequency representations with training them to insert probabilities into Bayes' rule. Frequency training gave a larger immediate learning effect and was more stable over time, while transfer to new problems was similar. Our post on common probability mistakes uses the same idea for a factory problem.
Which mistakes cost marks?
Reversing the conditional. P(result | cause) is the likelihood, not the answer.
Priors that do not add to 1. A missing cause gives a wrong denominator.
Stopping at the numerator. 9/30 is a branch, not the answer.
Changing the second bag wrongly in transfer problems. Rewrite its contents for each branch.
Skipping the check. Posteriors must lie between 0 and 1 and add to 1.
A practice routine
Do two problems of each type above: transfer, truth-telling, screening.
Solve each twice, once with the formula and once with counts of 1000, and confirm the answers match.
Draw the tree before writing any number.
Then move on to random variables and probability distributions, the next section of the chapter.
If you set a problem up and still get a different answer from the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, rework example 1 with 4 red and 1 blue ball in bag X, and check that your two posteriors add to 1.
