Independent Events
Two events are independent when one happening tells you nothing about the other, so P(A|B) = P(A). The test is multiplicative: A and B are independent exactly when P(A∩B) = P(A)·P(B). This is the rule for anything done "with replacement" or for genuinely separate trials — two coin tosses, two rolls of a die, drawing-and-returning a ball. Do not confuse independent with mutually exclusive: mutually exclusive events cannot occur together (overlap zero), whereas independent events usually can. For a chain of independent events, just multiply the individual probabilities. The CAT favourite is "at least one success in n trials", best solved by the complement: P(at least one) = 1 − P(no success)^n. For example, the chance of at least one six in two rolls of a die is 1 − (5/6)² = 11/36.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core probability rules
| Classical probability | P(E) = favourable / total |
|---|---|
| Range of probability | 0 ≤ P(E) ≤ 1 |
| Complement rule | P(E′) = 1 − P(E) |
| Addition rule | P(A∪B) = P(A) + P(B) − P(A∩B) |
| Mutually exclusive | P(A∩B) = 0 ⇒ P(A∪B) = P(A) + P(B) |
Conditional, independence & Bayes
| Conditional probability | P(A|B) = P(A∩B) / P(B) |
|---|---|
| Multiplication rule | P(A∩B) = P(B) × P(A|B) |
| Independent events | P(A∩B) = P(A) × P(B) |
| Total probability | P(A) = P(B)·P(A|B) + P(B′)·P(A|B′) |
| Bayes’ theorem | P(B|A) = P(B)·P(A|B) / P(A) |