Probabilities of an exhaustive set of outcomes sum to \(1\); multiply along branches of a tree (adjusting for 'without replacement'), add across mutually exclusive branches. Conditional probability is best handled with counts (two-way tables, Venn diagrams) rather than formulas.
Rules (M7.1-M7.7)
- \(P(\text{not } A) = 1 - P(A)\). \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\); for mutually exclusive events the last term is \(0\).
- Independent: \(P(A \text{ and } B) = P(A)P(B)\). Test independence by checking this product.
- Without replacement: the second probability uses the reduced totals (\(\frac{4}{10}\cdot\frac{6}{9} + \frac{6}{10}\cdot\frac{4}{9}\)).
- Conditional: \(P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}\) - with counts, restrict to the \(A\) group.
- Expected frequency \(= n \times p\). Relative frequency approaches the theoretical probability as trials increase, but any single run can differ.
Techniques
- Possibility spaces: two dice give \(36\) equally likely outcomes; 'product odd' needs both odd: \(\frac14\).
- 'At least one' = \(1 - P(\text{none})\). 'First success on the third go' \(= q^2 p\).
- Unknown probabilities: set up an equation, e.g. \(2p(1 - p) = \frac38\) gives a quadratic with two valid roots.
- Symmetry: with three fair coins, \(P(\text{more heads than tails}) = \frac12\).
Must-know (no formula booklet)
- \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\)
- independent: \(P(A)P(B)\)
- \(P(B|A) = \frac{P(A \text{ and } B)}{P(A)}\)
- at least one \(= 1 - P(\text{none})\)
Classic traps
- Mutually exclusive and independent are different: exclusive events with non-zero probabilities are never independent.
- Without replacement the denominators change on the second draw.
- 'Given that' fixes the denominator to the given group's size.
Worked example
Q. \(P(A) = 0.6\), \(P(B) = 0.5\), \(P(A \text{ and } B) = 0.3\). Independent?
A. \(0.6 \times 0.5 = 0.3\): yes. \(P(A \text{ or } B) = 0.8\).