Concepts: axioms and equally likely outcomes
Probability24 min readLesson 3 of 6
3.1 The classical definition
When a random experiment has finitely many outcomes and all of them are equally likely, the probability of an event is the number of outcomes favourable to the event divided by the total number of outcomes. A fair coin gives probability one half for each face, and a fair die gives probability one sixth for each number. The definition works only when the outcomes are symmetric in the sense that no outcome is more likely than another; a biased coin does not fit, and for such experiments probability is assigned through the axioms instead.
3.2 The axioms
The axiomatic approach does not depend on symmetry. Probability is a function on events satisfying three rules: every probability lies between 0 and 1, the probability of the whole sample space is 1, and the probabilities of mutually exclusive events add. All other rules follow from these three. Setting A equal to the whole sample space in the complement rule gives the probability of the empty set as 0. Writing the sample space as the union of an event and its complement, which are mutually exclusive, gives P(A) plus P(A complement) equal to 1, so the probability of not A is 1 minus the probability of A.
3.3 Worked examples: dice and cards
Two dice are rolled. There are 36 ordered outcomes, each with probability 1/36. The event the sum is 7 has six favourable outcomes, (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), so its probability is 1/6. The event the sum is 2 has only (1,1), so its probability is 1/36, and the event the sum is 12 has only (6,6), also 1/36. The sums near the middle of the range have more ways, which is why 7 is the most likely total. From a pack of 52 cards, the probability of an ace is 4/52, of a red card is 26/52, and of a spade is 13/52. These counts are worth memorising because they recur in many problems.
3.4 Counting equally likely outcomes
Most textbook problems reduce to counting. For a coin tossed n times there are 2 to the power n outcomes, and the number of sequences with exactly r heads is the binomial coefficient n choose r. For a die rolled twice there are 36 ordered pairs. For a hand of cards drawn from a pack, combinations count unordered hands while permutations count ordered selections, and the choice between them depends on whether the order matters. The same care is needed when objects are distinct: selecting two students from a class gives combinations, while awarding first and second prizes to two students gives permutations.
3.5 Empirical probability and its limits
Before the axioms were formalised, probability was estimated by repeating an experiment and counting. If an event occurs m times in n trials, the ratio m/n is the empirical probability, and it tends to settle near a fixed value as n grows. This is the frequency interpretation of probability. The classical definition agrees with it for symmetric experiments, but the classical definition cannot handle a biased coin or a loaded die, for which the frequencies themselves define the probability. The axiomatic approach covers both cases, because it starts from rules that any consistent assignment must satisfy rather than from symmetry.
3.6 Choosing between permutations and combinations
The counting method decides the answer, so identify what is being counted. If the problem selects a group without order, as when two students are chosen for a committee, use combinations. If it arranges the selected objects, as when prizes are awarded or letters form a code, use permutations. Drawing two cards together is a combination, while drawing one card, replacing it and drawing again is an ordered pair with 52 squared possibilities. The same distinction applies to coins and dice: the sequence HTH is different from THH even though both contain the same number of heads. Writing the sample space explicitly for small cases trains the habit of choosing correctly for larger ones.
3.7 Worked example: a die and a coin together
A fair coin is tossed and a fair die is rolled. The sample space has 12 ordered outcomes, each with probability 1/12. Let E be the event the coin shows a head and the die shows an even number. The favourable outcomes are H2, H4 and H6, so the probability is 3/12 = 1/4. Let F be the event the die shows a number greater than 1. Then F has 10 outcomes, so P(F) = 10/12 = 5/6. The complement of F is the single outcome in which the die shows 1, and its probability is 2/12 = 1/6, which is 1 minus 5/6 as the complement rule demands.
Tip
When a problem combines two experiments, multiply the sizes of the sample spaces to get the total. For a coin and a die together the total is 12, not 8 or 6.