Probability: conditioning, total probability and Bayes
Probability questions combine counting with conditional reasoning: describe the sample space, count cases with the right combinatorial tool, then apply conditional probability, the total probability theorem and Bayes' theorem, usually through a tree or a two-way table.
Core rules
- Laplace's rule P(A) = favourable/possible applies only when elementary outcomes are equiprobable.
- Counting: variations V(n,r) = n!/(n−r)! when order matters, permutations P(n) = n!, combinations C(n,r) = n!/(r!(n−r)!) when order does not; with replacement the counts change to n^r or C(n+r−1, r).
- P(A∪B) = P(A) + P(B) − P(A∩B); P(A^c) = 1 − P(A), the key to every "at least one" question.
- Conditional probability: P(A|B) = P(A∩B)/P(B). A and B are independent when P(A∩B) = P(A)P(B), equivalently P(A|B) = P(A).
- Total probability: if the A_i form a partition, P(B) = Σ P(A_i)P(B|A_i).
- Bayes: P(A_i|B) = P(A_i)P(B|A_i) / Σ P(A_j)P(B|A_j), used when you observe the result and update the probability of the source.
Worked example
Urn A holds 3 red and 2 blue balls, urn B holds 1 red and 4 blue. An urn is chosen at random and one ball is drawn; it turns out red. Here P(A) = P(B) = 1/2, P(R|A) = 3/5 and P(R|B) = 1/5. Total probability: P(R) = 1/2·3/5 + 1/2·1/5 = 3/10 + 1/10 = 2/5. Bayes: P(A|R) = (3/10)/(2/5) = 3/4. The tree makes the arithmetic transparent: multiply along the branches and add the branches that end in R.
Traps that cost marks
- Swapping P(A|B) and P(B|A).
- Treating mutually exclusive events as independent: if P(A∩B) = 0 and both have positive probability, they are dependent.
- Assuming independence without checking the product.
- Using combinations where order matters, or forgetting that draws are without replacement.
- In Bayes, forgetting to divide by the total probability of the observed event.