Classical Probability
Classical probability assumes every outcome is equally likely, so P(E) = (favourable outcomes)/(total outcomes). The whole skill is accurate counting. A single die has 6 outcomes; two dice have 36 ordered pairs; a standard pack has 52 cards (26 red, 26 black, 13 of each suit, 4 of each rank, 12 face cards). When you draw several items together, count with combinations: choosing r balls from n is C(n,r). The CAT-smart move for any "at least one" question is the complement: P(at least one) = 1 − P(none), which usually has a far shorter count. Also use the addition rule P(A∪B) = P(A) + P(B) − P(A∩B) for "A or B", subtracting the overlap so it is not counted twice. Always fix the sample space first — "with replacement" or "without replacement" silently changes the denominator.
✅ 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) |