Counting and Probability
The Counting and Probability domain tests whether you can count outcomes systematically and convert those counts into probabilities. Expect the fundamental counting principle, permutations versus combinations, arrangements with repeated items, probability rules for unions, complements, and conditional events, and expected value.
Counting methods
- Multiplication principle: if a first choice can be made in m ways and a second in n ways, the pair can be made in mn ways. Extend across all stages of a task.
- Permutations count ordered selections: nPr = n!/(n − r)!.
- Combinations count unordered selections: nCr = n!/(r!(n − r)!).
- Distinguishable arrangements of a multiset: n!/(n₁!·n₂!·…).
- Selections with repetition allowed: n^r outcomes.
Probability rules
- P(E) = favorable outcomes ÷ total outcomes, valid when outcomes are equally likely.
- Complement: P(not E) = 1 − P(E). "At least one" is almost always faster as 1 − P(none).
- Addition: P(A∪B) = P(A) + P(B) − P(A∩B); drop the subtraction only when the events are disjoint.
- Independence: P(A∩B) = P(A)·P(B). Conditional: P(A|B) = P(A∩B)/P(B).
- Expected value = Σ x·P(x).
Worked example
Nine students - five men and four women - apply for a three-person committee. How many committees are possible, and what is the probability that exactly one member is a woman? Committees are unordered, so use combinations: C(9,3) = 9!/(3!·6!) = 84. Exactly one woman means choosing 1 of 4 women and 2 of 5 men: C(4,1)·C(5,2) = 4·10 = 40. P = 40/84 = 10/21 ≈ 0.476.
Traps
- Using permutations when order is irrelevant inflates the count by a factor of r!.
- Two events connected by "or" are not disjoint unless stated; forgetting to subtract the overlap double-counts.
- Drawing without replacement changes the denominator on the second draw (4/9 then 3/8, not 4/9 twice).
- Expected value need not be an attainable outcome - a fair die's expected value is 3.5.