Probability and Bayes' Theorem
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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.