Probability rules & combinatorics
The idea
Counting is the engine of probability: when outcomes are equally likely, a probability is favorable outcomes over total outcomes, and both counts often demand combinatorics. The multiplication principle starts everything — a sequence of choices multiplies the number of options at each stage. Permutations count ordered arrangements; combinations count unordered selections, with C(n, k) = n!/(k!(n − k)!) ways to choose k items from n when order is irrelevant. The addition rule P(A or B) = P(A) + P(B) − P(A and B) and the complement rule P(not A) = 1 − P(A) round out the toolkit.
Ask one decisive question before any count: does order matter? Choosing a committee is a combination — the same four people in any order form the same committee — while electing a president and a treasurer is a permutation. For multi-condition counts, count each requirement separately and multiply. The misconception to dismantle is that 'or' always means plain addition; when events overlap, the outcomes in both get counted twice, and subtracting the overlap repairs the total. And when a problem says at least one, the complement is almost always the shorter road.
Worked example
A club has 6 women and 5 men. A committee of 4 people is chosen at random. What is the probability that the committee contains exactly 2 women and 2 men?
- Committees are unordered, so count with combinations. Total possible committees: C(11, 4) = (11 × 10 × 9 × 8)/(4 × 3 × 2 × 1) = 7920/24 = 330.
- Count the favorable committees in two stages: choose 2 women from 6 in C(6, 2) = 15 ways, and choose 2 men from 5 in C(5, 2) = 10 ways.
- By the multiplication principle, any pair of women can join any pair of men, so the favorable count is 15 × 10 = 150.
- Divide: P(exactly 2 women) = 150/330 = 5/11 ≈ 0.455. Plausibility: a 2–2 split is the most balanced outcome from a nearly even club, so it should be the likeliest single split yet still well below certainty — and 5/11 fits that profile.
Answer. The probability of exactly 2 women and 2 men is 5/11, about 0.455.
Check your understanding
- How do you decide whether a counting problem needs permutations or combinations, and what changes in the resulting formula?
- Why does the multiplication principle justify counting the women and the men separately and then multiplying?
- Why does P(A or B) require subtracting P(A and B), and in which situations can that subtraction be skipped safely?
- How would the complement rule shorten a question asking for the probability of at least one woman on the committee?
Build the foundations first
Probability rules & combinatorics builds on these concepts. If any feel shaky, start there.