When every outcome is equally likely, "how likely?" becomes nothing more than favorable ÷ total — so the art is in the counting.
Flip a fair coin a thousand times in Lesson 33.1 and the frequency of heads steadied near one number, the probability. That worked for any coin, fair or bent. But when the outcomes happen to be equally likely — the six faces of a fair die, the two sides of a fair coin, every shuffled deal of a deck — we do not need a thousand trials at all. We can simply count. List the outcomes you want, list all the outcomes there are, and divide. This is the classical model of probability, the oldest idea in the subject, and it turns every probability question into a counting question — exactly the permutations and combinations you mastered in Stage 32.
Roll a fair die. Nobody can name the face in advance — that is the randomness from Lesson 33.1. But there is extra structure here that a bent coin lacks: no face is favored over any other. The die is symmetric, so by symmetry each of the six faces must carry the same share of the likelihood. Six equal shares that must total 1, so each face has probability 1/6.
That symmetry is the whole engine. A model deserves the name classical when two things hold: the sample space Ω is finite, and every outcome in it is equally likely. A fair coin (two equal outcomes), a fair die (six), a well-shuffled deck (every deal equally likely), a lottery drum where every ball is identical — all qualify. The phrase to keep in mind is a level playing field: no outcome has an edge.
A classical probability model needs (1) a finite sample space Ω, and (2) every outcome equally likely. Then each of the N outcomes carries probability exactly 1/N.
Conclusion. When symmetry makes every outcome equally likely, we no longer estimate probability by experiment — each of N outcomes is worth exactly 1/N, and we are ready to count.
An event A is just a chunk of Ω — "the roll is even" is the chunk {2, 4, 6}. Because every single outcome is worth the same 1/N, the probability of A is simply how many outcomes it contains, each counted once:
P(A) = number of outcomes in Anumber of outcomes in Ω = favorabletotal
That is the classical formula. The numerator is amber because it is a count; the denominator is blue because it is the size of the event-space we measure against. Read "favorable over total" every time.
Roll a fair die; let A = "an even number." The favorable outcomes are {2, 4, 6}, so |A| = 3 and |Ω| = 6. Therefore P(A) = 3/6 = 1/2.
Toss two fair coins. The equally-likely outcomes are HH, HT, TH, TT — four of them, not three. Let A = "exactly one head" = {HT, TH}. Then P(A) = 2/4 = 1/2, while P(two heads) = 1/4. The four-outcome list — not a three-way "0, 1, or 2 heads" — is what makes the outcomes equally likely.
Now turn to the picture that says it all: the two-dice grid. Roll a red die and a blue die; each of the 6 × 6 = 36 ordered pairs is equally likely (the dice are distinguishable, so (2, 5) and (5, 2) are different outcomes). Pick an event below and watch the formula compute its probability by literally counting the lit cells — the helper returns the favorable and total counts, and we divide.
Conclusion. Once every outcome is equally likely, probability is bookkeeping: P(A) = favorable / total, with both numbers honestly counted — for the sum = 7 event that is 6/36 = 1/6.
"List all the outcomes" is fine for 36 dice-pairs, but hopeless for a deck of cards or a bag of marbles where the totals run into the millions. This is exactly where Stage 32 walks in: combinations are how we count favorable and total without writing the list. The classical fraction becomes a ratio of C(n, k)'s.
Here is the model case. A bag holds 3 red and 2 blue marbles. Reach in and grab 2 at once (order does not matter — it is a handful, not a sequence). What is the probability both are red?
The general shape, when a bag has M "wanted" items out of N total and you draw n of them, wanting exactly k wanted:
P = C(M, k) · C(N−M, n−k)C(N, n)
Choose your k wanted from the M wanted, choose the remaining n−k from the N−M unwanted, over all C(N, n) handfuls. (This is the hypergeometric count; it returns in full force in Lesson 33.8.) Dial the draw below and watch every count come straight from C(n, k).
When the list is too long to write, count it with combinations. P = favorable C'stotal C's — numerator and denominator are both C(n, k)'s, never the eye.
Conclusion. The classical formula and Stage 32 counting are the same tool: P(both red) = C(3, 2)/C(5, 2) = 3/10, and the combination form scales to any bag.
What if there are infinitely many equally-likely outcomes — a dart that can land anywhere on a board, a pebble dropped into any spot of a pond? You cannot count infinitely many points, so "favorable over total" has nothing to divide. But the spirit survives: replace count with area. If the landing point is equally likely to be anywhere in a region Ω, then the probability of landing in a target T is the area ratio:
P(T) = area of the targetarea of the whole region
This is geometric probability. The equally-likely assumption is still doing all the work — "uniformly at random over the region" is the area-world version of "every outcome equally likely." Drop a circular target into a square board: the chance a uniform dart hits the circle is just πr² over the square's area.
A square pond is 4 m on a side; a circular lily pad of radius 1 m floats at the center. A pebble dropped uniformly at random — where does it land? P(on the pad) = π·1²4² = π/16 ≈ 0.196. The probability is the area ratio, full stop.
Conclusion. When outcomes spread over a region instead of a list, swap counts for areas: P = target area / total area — geometric probability is the classical model in continuous clothing.
The classical formula is so easy that it is tempting to use it everywhere. But it has a licence condition: the outcomes you divide by must be equally likely. Forget that, and the fraction lies.
The famous cautionary tale: a family has two children; given that at least one is a boy, what is the probability both are boys? A careless count says: the families are BB, BG, GG — "both boys" is 1 of 3, so 1/3… and that is wrong, because BG is twice as likely as GG. Boy-then-girl and girl-then-boy are different, equally-likely outcomes. The honest sample space is the ordered list BB, BG, GB, GG — four equally-likely outcomes.
The classical formula P = favorable/total is only valid when the listed outcomes are equally likely. Writing two-child families as BB, BG, GG and concluding P(BB) = 1/3 is the classic trap — BG is twice as likely as GG. The real, equally-likely space is BB, BG, GB, GG, giving P(BB) = 1/4.
The same warning kills "sum of two dice has 11 outcomes 2…12, so each has probability 1/11." Those eleven sums are not equally likely — sum 7 has six ways, sum 2 has one. Always count the equally-likely elementary outcomes (the 36 pairs), never a coarser list.
Conclusion. Before you divide, check the licence: are the outcomes really equally likely? If not, go back to the finer, symmetric list — or you will quote 1/3 where the truth is 1/4.
| Tool | Formula | Use it when… |
|---|---|---|
| Classical probability | P(A) = favorable / total | Ω is finite and every outcome is equally likely |
| Counting the parts | C(M,k)·C(N−M,n−k) / C(N,n) | drawing a handful — count with combinations (Stage 32) |
| Geometric probability | P = target area / total area | a point lands uniformly over a region (infinitely many outcomes) |
| The licence check | are the outcomes equally likely? | always — BB/BG/GG ⇒ 1/3 is a trap; the truth is 1/4 |
Carry one sentence forward: classical probability is counting in disguise — equally-likely outcomes turn "how likely?" into favorable ÷ total, combinations do the counting, and area does it when the outcomes are a continuum. Next, in Lesson 33.3, events start to combine — "and," "or," "not" — and probability learns to add.
Six questions to lock it in. Tap the answer you think is right.
This lesson builds the classical (equally-likely) model of probability and computes probabilities by counting — directly addressing HSS-CP.A.1 (describe events as subsets of a sample space) and HSS-CP.B.9 (use permutations and combinations to compute probabilities of compound events). The two-dice grid, the draw-from-a-bag combination widget, and the area-ratio model each recompute an honest figure from a control, so students see that probability is a computed count, never a number read off a picture. The deliberate "two children" trap trains the habit every probabilist needs: before dividing favorable by total, verify the outcomes are genuinely equally likely. It connects forward to the addition rule (33.3) and back to combinations (Stage 32).