In short: To solve a Bayes' theorem problem, identify the partition of causes and their prior probabilities, write the likelihood P(evidence | each cause), find the total probability of the evidence, then divide the relevant joint probability by that total. In one line: P(cause|evidence) = P(evidence|cause)·P(cause) / P(evidence).
If you want to know how to solve Bayes theorem problems in JEE, the trick is to stop treating it as a formula to memorise and start treating it as a four-quantity bookkeeping exercise: priors, likelihoods, a total, and a final divide. JEE Advanced almost always frames Bayes as "an effect happened — which cause produced it?" Machines making defective parts, bags of balls, players of different skill. Get the structure right and every one of these problems collapses to the same five steps.
The reliable method
Apply these steps in order, every time:
- Identify the causes (the partition). List the mutually exclusive, exhaustive events that could produce the outcome — call them C₁, C₂, …. Their priors must sum to 1.
- Write the priors P(Cᵢ). These are the probabilities of each cause before you see the evidence.
- Write the likelihoods P(E|Cᵢ). For each cause, the probability of the observed evidence E given that cause.
- Compute the total probability P(E). Sum the products: P(E) = Σ P(E|Cᵢ)·P(Cᵢ). This is the denominator.
- Divide for the posterior. P(Cₖ|E) = P(E|Cₖ)·P(Cₖ) / P(E) for the cause k you care about.
A worked example
A workshop has two machines. Machine X makes 60% of all items and 2% of its output is defective. Machine Y makes the remaining 40% and 5% of its output is defective. An item is picked at random and found defective. What is the probability it came from Machine Y?
Step 1 — causes: X and Y (a partition; every item comes from exactly one).
Step 2 — priors: - P(X) = 0.60 - P(Y) = 0.40
Step 3 — likelihoods (defective given machine): - P(D|X) = 0.02 - P(D|Y) = 0.05
Step 4 — total probability of a defective item:
P(D) = P(D|X)·P(X) + P(D|Y)·P(Y) = (0.02)(0.60) + (0.05)(0.40) = 0.012 + 0.020 = 0.032
Step 5 — posterior for Machine Y:
P(Y|D) = P(D|Y)·P(Y) / P(D) = 0.020 / 0.032 = 5/8 = 0.625
So a defective item came from Machine Y with probability 5/8. The method holds because the partition X, Y covers every possibility exactly once, so the total probability correctly accounts for all the ways a defect can arise before Bayes redistributes that total to the cause you asked about.
Common mistakes to avoid
- Skipping the total-probability step. Fix: you cannot apply Bayes without P(E); build it from every cause first.
- Using a non-exhaustive partition. Fix: the causes must cover all possibilities and not overlap, so their priors sum to exactly 1.
- Swapping prior and likelihood. Fix: the prior is P(cause); the likelihood is P(evidence|cause). Keep them in their own rows.
- Dividing by the wrong denominator. Fix: always divide by the full P(E), not by a single prior or likelihood.
- Reporting P(E|Cₖ) as the answer. Fix: the question wants the reverse, P(Cₖ|E) — finish the division.
Frequently asked questions
What are the steps to apply Bayes' theorem in a JEE problem? List the causes and their priors, write the likelihood of the evidence under each cause, sum those products to get the total probability, then divide the chosen cause's product by that total.
What is the difference between the prior and the likelihood? The prior P(cause) is how likely a cause is before evidence. The likelihood P(evidence|cause) is how probable the evidence is assuming that cause. Bayes combines them.
Why do I need total probability inside Bayes' theorem? The denominator P(evidence) must account for every way the evidence can occur. Total probability sums those contributions across all causes so the posteriors are correctly normalised.
Do the priors always have to sum to 1? Yes. The causes form a partition of the sample space, so their prior probabilities are exhaustive and mutually exclusive, summing to exactly 1.
Can Bayes' theorem be used with more than two causes? Yes. Extend the total-probability sum over all causes C₁, C₂, …, Cₙ, then divide the relevant product by that sum. The five steps are identical.
Practise this
Find out if this trips you up — [take the free diagnostic](/diagnostic), then work through the [free Socratic lesson](/lesson/start) to drill the five-step method on fresh problems.