Probability and combinatorial counting
Probability and combinatorial counting
Probability supplies one of the statistics questions in the paper: typically a conditional probability or Bayes problem with urns, dice or diagnostic tests, or a counting problem leading to the binomial distribution.
Core rules
- Laplace's rule: probability equals favourable outcomes divided by possible outcomes, valid only when the outcomes are equally likely.
- Addition: P(A or B) = P(A) + P(B) - P(A and B); the last term disappears only when A and B are mutually exclusive.
- Conditional probability: P(A|B) = P(A and B)/P(B). Events A and B are independent exactly when P(A and B) = P(A)P(B). Independence and mutual exclusivity are different ideas.
- Total probability: if the events B1, ..., Bn partition the sample space, then P(A) is the sum of P(Bi)P(A|Bi). Bayes reverses the conditioning: P(Bi|A) = P(Bi)P(A|Bi) divided by that same sum.
- Counting: permutations n!/(n-r)! when order matters without repetition, combinations n!/(r!(n-r)!) when order is irrelevant, n^r sequences with repetition allowed, and n!/(n1!n2!...) arrangements of a multiset.
- Binomial: for X ~ B(n,p), P(X = k) = C(n,k) p^k (1-p)^(n-k), with mean np and variance np(1-p). For at least one success, use the complement.
Worked example
Urn A holds 3 red and 2 white balls; urn B holds 1 red and 4 white balls. An urn is chosen at random and a red ball is drawn. What is the probability that urn A was chosen?
By total probability, P(red) = (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 2/5. By Bayes, P(A|red) = (1/2)(3/5) divided by 2/5, that is (3/10)/(2/5) = 3/4.
Traps that cost marks
Adding probabilities without subtracting the intersection; confusing P(A|B) with P(B|A), the classic error in diagnostic-test problems; using combinations where order matters, or permutations where it does not; assuming independence because two probabilities look convenient; and forgetting that drawing without replacement changes p from one trial to the next.