# Probability and Bayes' Theorem

> Probability for JEE Mathematics; addition rules, conditional probability, Bayes theorem and binomial distribution with classic traps.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/probability
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Probability and Bayes' Theorem", PrepElephant, https://prepelephant.com/topics/jee/mathematics/probability

## Direct answer

P(A or B) = P(A) + P(B) - P(A and B), independence means P(A and B) = P(A) × P(B), and conditional probability is P(A given B) = P(A and B)/P(B). The law of total probability chains causes to an outcome — P(A) = sum of P(Ei) P(A given Ei) over a partition — and Bayes' theorem inverts the arrow: P(Ei given A) = P(Ei) P(A given Ei) divided by that same sum.

## What you must remember

- Addition rule: P(A union B) = P(A) + P(B) - P(A intersection B); mutually exclusive events drop the intersection term; complement rule P(not A) = 1 - P(A), the engine behind "at least one" = 1 - P(none).
- Conditional probability: P(A given B) = P(A intersection B)/P(B) for P(B) > 0; independent events satisfy P(A given B) = P(A), which is what independence means.
- Partition and total probability: if E1 to En are pairwise disjoint and exhaustive, P(A) = P(E1)P(A|E1) + ... + P(En)P(A|En).
- Bayes' theorem: P(Ei given A) = P(Ei)P(A given Ei)/[sum over k of P(Ek)P(A given Ek)] — reverse the conditioning with the observed outcome.
- Binomial distribution: for n independent trials with success probability p, P(exactly r successes) = C(n, r) p^r (1 - p)^(n - r); mean np, variance np(1 - p).
- Classical counting: probability = favourable equally likely cases / total cases; combinations do the counting, with or without replacement deciding the denominators.
- De Morgan in probability: P(not (A and B)) = P(not A or not B) and vice versa — complements convert awkward intersections into unions.

## Common confusion

Mutually exclusive is not independent — the two ideas pull in opposite directions. Disjoint events cannot occur together, so knowing one occurred makes the other impossible; independence requires the probability to be untouched by the information, which for non-trivial disjoint events fails automatically. The second trap is direction: P(disease given positive test) is not P(positive test given disease); draw the tree with causes first, compute the total probability of the observation, then apply Bayes to reverse.

## Exam-focused takeaway

JEE Main asks dice, card and ball problems, conditional probability, direct Bayes with two or three causes and binomial-distribution numericals — abundant, formula-driven marks. JEE Advanced layers sequential experiments, at-least-one constructions through complements, probability fused with combinatorial identities, and conditional problems where the sample space quietly shrinks. The scoring habit: name the events in words, fix the partition, write the tree, and only then reach for formulas.

## Frequently asked questions

### What is Bayes' theorem used for?

Reversing conditional probability — from the likelihood of the observation under each cause to the probability of each cause given the observation, divided by the total probability of the observation.

### Can two events be both mutually exclusive and independent?

Only in the trivial case where one has probability zero; otherwise disjointness forces dependence, since one occurring rules the other out.

### What are the mean and variance of a binomial distribution?

np and np(1 - p) for n independent trials with success probability p; both come straight off the distribution, with no extra conditions.

### How do I compute "at least one" probabilities?

Through the complement: 1 - P(none) — almost always faster than summing the individual cases, especially with independence.

### What is a partition of the sample space?

A collection of pairwise disjoint events whose union is everything; total probability runs over a partition, and Bayes' posterior probabilities over one sum to 1.