When events ignore each other you multiply; when one event is given, you shrink the stage — and out of that fall the multiplication rule, total probability, and Bayes.
Yesterday's coin does not remember today's. Flip a fair coin, get heads, flip again — the second flip is still 0.5 for heads, untouched. Events that ignore each other like this are independent, and for them probability simply multiplies. But most interesting questions are the opposite: a clue arrives, and it changes the odds. "It's an even roll — now, is it a six?" The word now shrinks the whole sample space down to the even faces, and the probability we read off that smaller stage is a conditional probability. From these two ideas — multiply when independent, shrink when conditioning — grows the entire machinery of this lesson: the multiplication rule for drawing without replacement, the total-probability formula that mixes several sources, and a first look at Bayes, which runs the tree backward to ask "given the result, where did it come from?"
Two events A and B are independent when knowing that one happened tells you nothing about the other. A coin flip and a die roll; today's weather and the lottery; the first card if you replace it before the second. The test is an equation:
A, B independent ⟺ P(A∩B) = P(A)·P(B).
Why a product? Picture a tree. The first flip splits into H and T; from each of those, the second flip splits again into the same branch probabilities, because the second flip doesn't care what the first did. The chance of a particular path — say H then H — is the chance of the first branch times the chance of the second: 0.5·0.5 = 0.25. Independence is exactly the statement "the branch probabilities are the same at every node."
Let A = "first flip is heads," B = "second flip is heads." Each has probability 0.5. They are independent, so
P(A∩B) = P(HH) = 0.5·0.5 = 0.25.
All four paths HH, HT, TH, TT come out to 0.25 each, and 0.25·4 = 1 — the leaves of an honest tree always sum to 1.
Conclusion. Independent events multiply: P(A∩B) = P(A)·P(B), and a tree with the same branch probabilities at every node is exactly that picture.
Now suppose someone peeks and tells you a partial fact. Roll one fair die. Before you look, P(six) = 1/6. But you're told A = "the roll is even." That sentence does real work: it throws away the odd faces and leaves you on a smaller stage, {2, 4, 6}. Inside that restricted space, a six is now one of three equally likely faces, so the chance of a six given even is 1/3. We write it P(B|A) — "the probability of B given A."
The definition makes "shrink the stage" precise. Restrict to A, then ask what fraction of that world is also in B:
P(B|A) = P(A∩B)P(A) (provided P(A) > 0).
The denominator P(A) is the size of the new stage; the numerator P(A∩B) is how much of B survived inside it. For the die: P(even ∩ six) = 1/6, P(even) = 1/2, so P(six | even) = (1/6)/(1/2) = 1/3 — matching the count.
Conclusion. Conditioning shrinks the sample space to A, and P(B|A) = P(A∩B) / P(A) reads the chance of B off that smaller stage.
Rearrange the conditional definition and you get a tool for building a probability one stage at a time. Multiply both sides by P(A):
P(A∩B) = P(A)·P(B|A).
Read it left to right as a story: first A happens (probability P(A)), and then, given that, B happens (probability P(B|A)). This is the natural rule for drawing without replacement, where the first draw changes the second's odds. Independence is just the special case where the clue doesn't matter, P(B|A) = P(B), and the rule collapses back to P(A)·P(B).
A bag holds 5 red and 3 blue marbles (8 total). Draw two without replacing the first. What is P(both red)?
Stage 1: P(first red) = 58. Stage 2, given a red is gone, only 4 reds remain among 7: P(second red | first red) = 47.
P(both red) = 5/8·4/7 = 20/56 = 5/14 ≈ 0.357.
The trap: writing 5/8·5/8 as if the draws were independent. They are not — pulling a red marble really does lower the next red's chance.
Conclusion. The multiplication rule chains two stages: P(A∩B) = P(A)·P(B|A) — multiply the first branch by the conditional second branch, exactly as a tree path does.
Often the thing you care about can arrive by several different routes, and you must weight each route by how likely it is. Suppose a white ball could come from box 1, box 2, or box 3, and the boxes have different white-ball rates. The boxes A₁, A₂, A₃ split the sample space — they are disjoint and together cover everything (a partition). Then the chance of a white ball is the sum, over boxes, of "pick that box, then draw white from it":
P(B) = P(A₁)·P(B|A₁) + P(A₂)·P(B|A₂) + P(A₃)·P(B|A₃) = Σ P(Aᵢ)P(B|Aᵢ).
On a tree, this is nothing more than "add the leaves where B happens." Each white leaf is the product of its box-edge and its white-edge; total up the white leaves and you have P(B). The widget below makes every one of those numbers visible.
Box 1 is 1/2 white, Box 2 is 1/5 white, Box 3 is 4/5 white; you pick a box at random (each 1/3), then a ball.
P(white) = ⅓·½ + ⅓·⅕ + ⅓·⅘ = 1/6 + 1/15 + 4/15 = 0.5.
Each term is a white leaf of the tree; P(white) is their sum.
Conclusion. When an event can arrive by a partition of routes, weight each route by its share: P(B) = Σ P(Aᵢ)P(B|Aᵢ) — the sum of the matching leaves.
The tree we just built runs forward: box first, then color. Bayes asks the backward question. You see a white ball — which box did it most likely come from? That's P(Aᵢ|B), the conditional probability the other way around. By the definition of conditional probability and the multiplication rule,
P(Aᵢ|B) = P(Aᵢ)·P(B|Aᵢ)P(B) = one white leafall white leaves.
The numerator is the single white leaf for box i; the denominator is the total-probability sum from §33.4.4. Bayes is literally one leaf divided by the sum of the leaves of the same color. Use the widget's Backward toggle to see it: pick a box, and the readout shows that box's white leaf over the white total.
With the boxes above, the white leaves are 1/6 ≈ 0.167 (box 1), 1/15 ≈ 0.067 (box 2), 4/15 ≈ 0.267 (box 3), summing to P(white) = 0.5. So
P(box 3 | white) = 4/151/2 = 8/15 ≈ 0.533.
A white ball points most strongly at box 3 — the box with the highest white rate — exactly as intuition wants.
1. Mutually exclusive ≠ independent. If A and B are disjoint (can't both happen, P(A∩B)=0) and each has positive probability, then learning A happened forces B to 0 — so they are about as dependent as two events can be. Disjoint is the opposite of independent, not the same thing.
2. P(B|A) ≠ P(A|B) — the base-rate trap. "Most accidents involve drivers who had breakfast" does not mean breakfast causes crashes; the two conditionals run in opposite directions and usually differ. P(white | box 3) = 4/5, but P(box 3 | white) = 8/15 — same two events, very different numbers. Bayes is the bridge between them, not a license to swap them.
Conclusion. Bayes reverses the tree: P(Aᵢ|B) = P(Aᵢ)P(B|Aᵢ) / P(B) — one leaf over the sum of same-colored leaves, and never confuse it with P(B|Aᵢ).
| Idea | Formula | Picture / when |
|---|---|---|
| Independent events | P(A∩B) = P(A)·P(B) | same branch probabilities at every node |
| Conditional probability | P(B|A) = P(A∩B)/P(A) | shrink Ω to A; read B off the smaller stage |
| Multiplication rule | P(A∩B) = P(A)·P(B|A) | chain two stages; drawing without replacement |
| Total probability | P(B) = Σ P(Aᵢ)P(B|Aᵢ) | sum the matching leaves over a partition |
| Bayes | P(Aᵢ|B) = P(Aᵢ)P(B|Aᵢ)/P(B) | one leaf ÷ the sum; reverse the tree |
| Two traps | disjoint ⇒ dependent | P(B|A) ≠ P(A|B) — don't swap conditionals |
Counting the leaves you sum is the classical model again; the events themselves are sets you intersect and union. Next, when a count is too hard to do by hand, we'll simulate instead.
Independent — the second flip can't know the first. P(both heads) = 0.5·0.5 = 0.25.
A = {1,3,5}, so P(A) = 3/6 = 1/2. A∩B = {3}, so P(A∩B) = 1/6. Then P(B|A) = (1/6)/(1/2) = 1/3. (Check by counting: of the three odd faces, only 3 is a multiple of 3 — 1 of 3.)
Multiplication rule. P(first good) = 4/6 = 2/3; given that, 3 good remain of 5, so P(second good | first good) = 3/5. P(both good) = 2/3·3/5 = 6/15 = 2/5 = 0.4. Drawing 4/6·4/6 would be the independence trap.
Total probability over the two machines. P(defective) = 0.60·0.02 + 0.40·0.05 = 0.012 + 0.020 = 0.032 (3.2%). Each term is a "defective" leaf of the tree.
Bayes: one leaf ÷ the sum. P(II | defective) = 0.40·0.050.032 = 0.020/0.032 = 0.625. Even though Machine II makes fewer parts, its higher defect rate makes it the more likely source of a bad one.
Mutually exclusive means A∩B = ∅, so P(A∩B) = 0. Then P(B|A) = 0/0.3 = 0. For independence we'd need P(A∩B) = P(A)P(B) = 0.3·0.4 = 0.12 ≠ 0, so they are not independent — they are dependent. Disjoint ≠ independent.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the core of the Common Core conditional-probability cluster. It interprets independence as HSS-CP.A.2 — A and B are independent exactly when P(A∩B)=P(A)P(B). Conditional probability and its interpretation in context are HSS-CP.A.3 and HSS-CP.B.6 (P(B|A) as the fraction of A's outcomes that also lie in B), and the multiplication rule P(A∩B)=P(A)P(B|A) is HSS-CP.B.8. Total probability and the first look at Bayes extend these to a partition of the sample space and the reversal of a probability tree. The two flagged misconceptions — confusing "mutually exclusive" with "independent," and swapping P(B|A) with P(A|B) (the base-rate trap) — are the errors students most reliably make, so they are worth naming aloud.