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Module 2 · Probability Foundations · PR-1
The position piece runs, and one reply sticks: "Six short nights in a row. Statistically, tonight has to be a good one — I'm due." A Toronto casino once banned a player for betting $200,000 on exactly that logic. Before Maya can answer either of them, she needs the mathematics of chance — starting from zero.
Live session · join with code OTTO-PR1 · ottolearn.org/join
Same rules as always: vote before I explain, every poll is anonymous, and changing your mind after discussion is the goal.
Today · the reply Maya owes her reader
Sample spaces and events
Classical, empirical, subjective probability
The axioms and the complement rule
Unions, intersections, the addition rule
Mutually exclusive vs. independent
Five questions, one mailbag reply. Every one starts with your vote.
Recall · last class · 45 seconds
Warm-up · 60 seconds · anonymous
Hold your vote. Today is about where probability numbers legitimately come from — and the counting shortcut has fine print.
Core concept 1 · C1
Before any number: list the outcomes. An experiment is any process whose outcome is uncertain — a die roll, a coin flip, spin #10 of a roulette wheel.
The set of every outcome the experiment could produce. Roll a die: \(S = \{1, 2, 3, 4, 5, 6\}\). Nothing outside the list can happen.
\(S\)Any collection of outcomes you care about — a subset of \(S\). "Roll an even number" is the event \(A = \{2, 4, 6\}\).
\(A \subseteq S\)The die shows 4 → the outcome landed inside \(A\), so \(A\) occurred. One roll can make several events occur at once: 4 is even and greater than 3.
Every tool today is built on this picture — outcomes in a set, events as subsets. Get the sample space right and the rest is counting.
Core concept 2 · C2
When every outcome in \(S\) is equally likely, probability is a counting ratio: outcomes in \(A\) over outcomes in \(S\). Even number on a fair die: \(P(A) = 3/6 = 1/2\).
\(P(A) = \dfrac{|A|}{|S|}\)One condition: equally likely. Fair dice, fair coins, shuffled decks qualify. A bent coin still has \(S = \{H, T\}\) — but \(P(H) \neq 1/2\). The formula fails the moment the outcomes aren't uniform.
"Rain or no rain — 50%" applies the counting formula to outcomes that are not equally likely. Two outcomes never means \(1/2\) — that's the fine print doing its job.
When the outcomes aren't equally likely, counting is worthless — you need data. That's the next tool, and it's the one the reader's streak needs.
Practice · classical probability · fresh numbers on demand
Commit first · the reader's logic, tested
Core concept 3 · C3
No equal-likelihood to lean on? Run the experiment \(n\) times, count how often \(A\) happened (\(f\)), and estimate the probability by the relative frequency — the same ratio you've computed since the first data class.
\(P(A) \approx \dfrac{f}{n}\)As \(n\) grows, \(f/n\) settles toward the true probability. It converges by averaging over ever more trials — not by tails "catching up." No debt is ever repaid.
Of her 300 sleep responses, 81 reported under 6 hours. Last month that ratio described her data; today it predicts: pick a respondent at random and \(P(\text{under 6 h}) = 0.27\).
\(\tfrac{81}{300} = 0.27\)Watch both claims at once: the proportion converges — and the heads-tails gap does not shrink. Convergence comes from averaging, not from evening out.
See it · C3
Predict first: after a long run of tails, does the running proportion snap back? Flip in bursts, watch the streak strip — and notice the predictor never budges from \(0.50\).
Click a button to start flipping.
Core concept 4 · C4 · enrichment
Some events can't be repeated: this start-up succeeding, this election, this exam. A number can still express an informed degree of belief — "I'd put it at 30%."
A subjective probability must still obey every rule of probability: between 0 and 1, complements summing to 1. A gut feeling that breaks the axioms isn't a probability — it's a mistake with confidence.
Counting (classical), data (empirical), judgment (subjective) — every probability you'll ever meet comes from one of the three. Ask: could I repeat this? Are the outcomes equally likely?
For the curious: classifying the type is context, not an exam skill — but it sharpens how you read every "70% chance" in the news.
Try it · C4
Decide the type before flipping each card. Where does Maya's 0.27 sit? And the forum's "50% rain"?
Pick the type for each scenario below.
Read the scenario, then pick the type of probability it is:
Core concept 5 · C5
Every probability lives between 0 (impossible) and 1 (certain). A claimed probability of 1.05 — or −0.2 — is broken on arrival, whatever its source.
\(0 \leq P(A) \leq 1\)The full sample space has probability exactly 1. This is what makes complements work: an event and its opposite split the whole 1 between them.
\(P(S) = 1\)The empty event — no outcome at all — has probability 0. And when two events can't overlap, their "or" is a plain sum. (The fine print on that arrives in a few slides.)
\(P(\emptyset) = 0\)The axioms are the fraud detector: any claimed set of probabilities that breaks them is wrong before you check anything else.
See it · C5
Predict first: push \(P(A)\) and \(P(B)\) until their raw sum passes 1 — which axiom, if any, breaks? Then drag the overlap and watch the union needle pull back inside the valid zone.
Quick check · spot the broken claim
Core concept 6 · C6
Everything not in \(A\) is the complement \(A'\). Together they fill \(S\) and never overlap — so they split 1 between them: whatever \(A\) doesn't take, \(A'\) gets.
\(P(A') = 1 - P(A)\)When an event is hard to count directly, count what it isn't. "At least one…" questions almost always fall to this: compute the probability of none, subtract from 1.
The move is a decision, not just a formula: spot the complement, then subtract instead of recounting. Watch me make that call out loud:
I do · watch me choose the shortcut
Full support — watch me decide to flip to the complement, then subtract.
The forecast gives \(P(\text{rain}) = 0.35\); Maya needs \(P(\text{no rain})\) for a survey-day plan. "Rain" and "no rain" fill every outcome and can't overlap — textbook complements.
Rather than tally every dry scenario, I take the whole \(1\) and remove the slice I don't want: \(1 - 0.35\).
\(= 0.65\). Check: it sits in \([0,1]\) ✓, and \(0.35 + 0.65 = 1\) ✓. Bigger than the rain chance — which fits a mostly-dry forecast.
The habit: name the complement, subtract from 1, check the sum. Now Maya's own survey number — you take the last step.
We do · you finish it
Support fades — the setup is given; you supply the subtraction and the warning.
From the survey: \(P(\text{a random respondent sleeps under 6 h}) = 0.27\).
So \(P(\text{6 h or more}) = ?\) — and what should Maya warn the reader this number is not?
Practice · the complement rule · fresh numbers on demand
On your own now — fresh numbers, every step yours. Re-roll as many as the room needs.
Core concept 7 · C7
\(A \cup B\): at least one of the two happens — \(A\), \(B\), or both. In a two-way table the union covers three cells: each "only" cell plus the overlap.
\(A \cup B\)\(A \cap B\): both happen at once. In the table it is a single cell — the one where the row and the column cross.
\(A \cap B\)You've been reading these since the first data class — every cell of a two-way table is an intersection, every row total a marginal. Today the counts become probabilities.
See it · C7
Predict first: click the top-left cell — which Venn region lights up? Then flip to probabilities and find the cell whose value is \(P(A \cap B)\).
Survey of 120 students — click any cell
| Passed B | Failed B′ | Total | |
|---|---|---|---|
| Groups A | 48 | 12 | 60 |
| Alone A′ | 42 | 18 | 60 |
| Total | 90 | 30 | 120 |
Venn diagram (n = 120)
Click any highlighted cell or total to see the corresponding Venn region.
Which number? · mixes DS-1 + PR-1
Core concept 8 · C8
Add \(P(A) + P(B)\) and every outcome in the overlap is counted twice — once inside each event. The raw sum can even sail past 1, as the thermometer just showed.
Subtract the overlap once and the double-count is exactly undone. This works for any two events — overlapping or not.
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)Watch the decision, not just the formula. The expert move is a check that happens before any arithmetic:
See it · C8
Predict first: with the overlap at zero, how do \(P(A) + P(B)\) and \(P(A \cup B)\) compare? Then grow the overlap and watch the two pull apart by exactly \(P(A \cap B)\).
Each dot is 1 of 100 students. A = plays a sport, B = is in a club.
I do · watch the whole decision
Full support — watch me make every decision, especially the overlap check.
Draw one card from a fair 52-card deck. \(A\) = heart, \(B\) = face card (J, Q, K). Fair deck → equally likely → classical: \(P(A) = \tfrac{13}{52}\), \(P(B) = \tfrac{12}{52}\).
Before I add anything: can one card be both? Yes — J♥, Q♥, K♥. So \(P(A \cap B) = \tfrac{3}{52} \neq 0\), and plain adding would count those three cards twice.
\(P(A \cup B) = \tfrac{13}{52} + \tfrac{12}{52} - \tfrac{3}{52} = \tfrac{22}{52}\). Subtracting once removes the second copy of each overlap card — they stay counted exactly once.
\(\tfrac{22}{52} \approx 0.42\): on the 0–1 scale, larger than either event alone, smaller than the raw sum \(\tfrac{25}{52}\). Exactly what a repaired double-count should look like.
The habit: overlap first, arithmetic second. Now Maya's own table — you take the last step.
We do · you finish it
Support fades — the setup is done for you; you commit to the final step.
Same crosstab as the poll: \(P(\text{part-time}) = 0.40\), \(P(\text{under 6 h}) = 0.27\), \(P(\text{both}) = 0.15\).
Finish it: \(P(\text{part-time OR under 6 h}) = ?\)
Practice · the General Addition Rule · fresh numbers on demand
On your own now — fresh numbers, every step yours. Re-roll as many as the room needs.
Core concept 9 · C9
Two events are mutually exclusive when they cannot both happen on the same trial — the overlap is empty. One die roll can't show both a 1 and a 6.
\(P(A \cap B) = 0\)Nothing to double-count → the addition rule sheds its correction: \(P(A \cup B) = P(A) + P(B)\). A special case — the general rule still holds; the subtraction is just \(-\,0\).
"They're about different things" is not a proof. "Even" and "greater than 4" sound unrelated — yet 6 is both. The only test is the sample space: is \(P(A \cap B) = 0\)?
A student ran the shortcut without the check. One line of the work below is wrong — find it before the votes reveal it.
Find the flaw · one line is wrong
Commit first · the semester's stickiest trap
Core concept 10 · C10
A statement about the sample space: the two events share no outcomes. In the Venn picture, the circles don't touch.
\(P(A \cap B) = 0\)A statement about knowledge: learning that \(B\) occurred leaves the probability of \(A\) untouched. The coin's five tails told us nothing about flip #6 — that's independence at work. (Made precise next class.)
\(P(A \mid B) = P(A)\)If both events have positive probability and can't co-occur, one happening slams the other to 0. Mutually exclusive events are never independent — they are maximally dependent.
The reader's fallacy and this trap are cousins: both invent a relationship ("due") or deny one ("separate") instead of checking what the sample space actually says.
See it · C10
Predict first: in the mutually-exclusive panel, what happens to \(P(B)\) the instant \(A\) occurs? Then run the independent panel and compare.
Mutually Exclusive
Die roll — A = {1, 2}, B = {4, 5, 6}
Independent
Two coins — A = Flip 1 = H, B = Flip 2 = H
Practice · mutually exclusive or not? · fresh scenarios on demand
Production task · write the reply
Exit ticket · 60 seconds · anonymous
The reply Maya files
The wheel has no memory: a streak changes nothing about the next trial. Probability is a property of the process, not a debt the past owes you.
A valid probability is a share of the sample space: between 0 and 1, complements splitting 1, overlaps subtracted exactly once.
"Can't both happen" is the opposite of "unrelated" — mutually exclusive events are maximally dependent.
Next class, the reader's follow-up: "But my nights aren't coin flips — doesn't one bad night cause the next?" Fair question. That's conditional probability, and it gets its own machinery.
Vote counts on today's slides are simulated for rehearsal; live sessions show the room's real votes.