Probability

Probability rewards clean sample-space thinking and the conditional/Bayes machinery. Advanced favours conditional probability, total probability and Bayes' theorem, independence, and binomial/geometric distributions. Careful modelling and the addition/multiplication rules do most of the work.

Key results

• Basics: P(A) = favourable/total (equally likely); 0 ≤ P ≤ 1; P(A′) = 1 − P(A).
• Addition: P(A∪B) = P(A) + P(B) − P(A∩B); mutually exclusive → no overlap.
• Conditional/independence: P(A|B) = P(A∩B)/P(B); independent iff P(A∩B) = P(A)P(B).
• Total probability & Bayes: P(A) = ΣP(A|Bi)P(Bi); P(Bi|A) = P(A|Bi)P(Bi)/ΣP(A|Bj)P(Bj).
• Binomial: P(k successes) = nCk p^k q^(n−k); mean np, variance npq.

Rule	Formula
Union	P(A)+P(B)−P(A∩B)
Conditional	P(A∩B)/P(B)
Bayes	P(A
Binomial P(k)	nCk p^k q^(n−k)

Exam Tricks & Tips

• 🎯 Bayes' theorem updates a prior given evidence: use it whenever the question asks "given the outcome, which cause?" — identify the partition Bi first.
• 🎯 Independence means P(A∩B) = P(A)P(B), which is not the same as mutual exclusivity (mutually exclusive events with nonzero probability are dependent).
• 🎯 Use the complement for "at least one" — P(at least one) = 1 − P(none) — far faster than summing cases.
• 🎯 Binomial applies for fixed n independent trials with constant p; mean np and variance npq are quick to quote.
• 🎯 Draw a tree for multi-stage experiments and multiply along branches, then add across paths (total probability).
• ❌ Common mistake: confusing P(A|B) with P(B|A) — Bayes exists precisely because these differ; keep the conditioning direction straight.

Expected exam pattern

2 questions likely: a conditional/Bayes problem and a binomial-distribution or combined-event calculation (often integer or multiple-correct). Modelling-heavy.

Quick recap

P(A∪B) = P(A)+P(B)−P(A∩B); conditional P(A|B)=P(A∩B)/P(B); independent iff product rule holds. Total probability and Bayes for cause-given-effect. "At least one" via complement. Binomial nCk p^k q^(n−k), mean np, variance npq.

Cover the answer, recall, then check. 12 cards on probability for JEE Advanced.

Q1. Addition rule for two events.
A1. P(A∪B) = P(A) + P(B) − P(A∩B).

Q2. Definition of conditional probability.
A2. P(A|B) = P(A∩B)/P(B).

Q3. Condition for two events to be independent.
A3. P(A∩B) = P(A)·P(B).

Q4. Bayes' theorem.
A4. P(Bi|A) = P(A|Bi)P(Bi) / Σ P(A|Bj)P(Bj).

Q5. Total probability theorem.
A5. P(A) = Σ P(A|Bi)P(Bi) over a partition {Bi}.

Q6. Fastest way to compute "at least one" probability.
A6. 1 − P(none) (complement).

Q7. Binomial probability of exactly k successes.
A7. nCk p^k q^(n−k), q = 1 − p.

Q8. Mean and variance of a binomial distribution.
A8. Mean = np; variance = npq.

Q9. Are mutually exclusive events with nonzero probabilities independent?
A9. No — if they cannot co-occur, P(A∩B) = 0 ≠ P(A)P(B).

Q10. Probability of the complement of A.
A10. P(A′) = 1 − P(A).

Q11. Difference between P(A|B) and P(B|A).
A11. They condition on different events and are generally unequal (related by Bayes).

Q12. When does the binomial model apply?
A12. Fixed number of independent trials, each with the same success probability p.

Probability rewards careful modelling of the sample space and fluent use of conditional reasoning. JEE Advanced favours Bayes' theorem, independence subtleties, and binomial/expected-value problems — where a clear event definition prevents most errors.

Core concept: probability is favourable/total outcomes for equally likely cases, extended by the addition and multiplication rules, and inverted by Bayes' theorem for "given the effect, find the cause".

Core ideas

Beginner — axioms and basic rules

P(A) ∈ [0, 1]; P(A∪B) = P(A) + P(B) − P(A∩B) (addition rule); P(A') = 1 − P(A) (complement — often the easy route for "at least one"). For mutually exclusive events, P(A∩B) = 0. Model the sample space carefully (ordered vs unordered, with/without replacement) — most mistakes are modelling errors, not arithmetic.

Intermediate — conditional probability and independence

Conditional: P(A|B) = P(A∩B)/P(B). Multiplication rule: P(A∩B) = P(A)P(B|A). Events A, B are independent iff P(A∩B) = P(A)P(B) — distinct from mutually exclusive (which are dependent unless one has probability 0). Pairwise independence does not imply mutual independence — a classic JEE trap.

Advanced — total probability, Bayes, and distributions

Total probability: if B₁…Bₙ partition the space, P(A) = Σ P(Bᵢ)P(A|Bᵢ). Bayes' theorem inverts this: P(Bₖ|A) = P(Bₖ)P(A|Bₖ) / Σ P(Bᵢ)P(A|Bᵢ) — the "given a defect, which machine made it" workhorse. Binomial distribution: n independent trials, success probability p, P(X = r) = ⁿCᵣ pʳ(1−p)^(n−r), mean np, variance np(1−p). Expected value E(X) = Σ x·P(x) is the tool for "expected number/gain" problems. Random-variable expectation and its linearity (E(X+Y) = E(X)+E(Y), even for dependent variables) simplify many otherwise-hard counts.

Worked example

A factory has two machines: A makes 60% of items with a 2% defect rate, B makes 40% with a 5% defect rate. An item is found defective — what is the probability it came from B? By Bayes: P(B|defective) = P(B)P(def|B) / [P(A)P(def|A) + P(B)P(def|B)] = (0.4×0.05) / (0.6×0.02 + 0.4×0.05) = 0.02 / (0.012 + 0.02) = 0.02/0.032 = 0.625. So there is a 62.5% chance the defective item came from B — even though B makes fewer items, its higher defect rate makes it the more likely source. This "reverse the conditioning" is the archetypal Bayes question.

How JEE Advanced tests this

Bayes' theorem "which source" problems; conditional probability and independence (including the pairwise-vs-mutual distinction); binomial-distribution "exactly/at least k successes" and mean/variance; expected value and linearity of expectation; "at least one" via complement; and geometric/uniform continuous probability (favourable region ÷ total).

Exam tricks & shortcuts

• "At least one" ⇒ 1 − P(none) is almost always easier.
• Independent: P(A∩B) = P(A)P(B); this is not the same as mutually exclusive.
• Bayes: P(cause|effect) = P(cause)P(effect|cause)/P(effect).
• Binomial mean np, variance np(1−p); use E linearity for expected counts.

Key formulas

• P(A) = favourable/total; P(A∪B) = P(A)+P(B)−P(A∩B).
• Conditional: P(A|B) = P(A∩B)/P(B); independent P(A∩B) = P(A)P(B).
• Total probability: P(A) = ΣP(Eᵢ)P(A|Eᵢ).
• Bayes: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/ΣP(Eⱼ)P(A|Eⱼ).
• Binomial: P(X=r) = ⁿCᵣ pʳq^(n−r); mean np, variance npq.

Use conditional probability and Bayes' theorem for dependent events; the binomial law models repeated independent trials.

Worked Example

Problem: In a factory, machines A, B, and C produce 25%, 35%, and 40% of the total output, with defect rates of 5%, 4%, and 2% respectively. An item is picked at random and found to be defective. Find the probability that it was produced by machine C.

Solution:
Prior probabilities: P(A) = 0.25, P(B) = 0.35, P(C) = 0.40.
Likelihoods of a defect: P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02.

Total probability of a defective item (law of total probability):
P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C)
= (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02)
= 0.0125 + 0.0140 + 0.0080 = 0.0345.

By Bayes' theorem:
P(C|D) = P(C)P(D|C) / P(D) = 0.0080 / 0.0345 ≈ 0.232.

Answer: P(the defective item came from machine C) ≈ 0.232 (about 23.2%).