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Module 3 · Probability Distributions · PR-5
Two classes ago Maya counted: 35 of the 128 diary patterns have three short nights — about 27%, if every pattern were equally likely. Last class she learned to weight outcomes instead of assuming they're equal. Today the two moves fuse into one formula: counting × weighting = the binomial distribution — and the reader's diary finally gets its honest answer.
Live session · join with code OTTO-PR5 · 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 · one formula, earned four ways
The four BINS conditions
The binomial formula — counting × weighting
Cumulative probabilities and the ±1 boundary
Mean \(np\), SD \(\sqrt{np(1-p)}\), and the shape of it all
Non-binomial settings and their proper models
Check the conditions, build the formula, stretch it to "at least", read off its center and spread — and know when to refuse it. Every step starts with your vote.
Recall · last class · 45 seconds
Warm-up · 60 seconds · anonymous
One specific pattern is just the multiplication rule from the start of the module. The hard question — and today's whole point — is "exactly \(k\) short nights in any order." That needs counting too.
Core concept 1 · C1
Today's shortcut is powerful and picky. It applies only when the situation passes four checks — remembered as BINS. Checking them is always step 1, the way a journalist verifies a source before quoting it.
Each trial ends exactly one of two ways — "success" or "failure." The labels are arbitrary: a short night is the success here, because it's what we're counting.
No trial's outcome changes any other trial's probability. This is the assumption the reader herself disputes — "one bad night causes the next" — and Maya must own it in print.
The number of trials \(n\) is set in advance: a week has 7 nights before any of them happens. "Keep going until…" settings have no fixed \(n\) — and fail this check.
The success probability \(p\) is identical on every trial — \(0.4\) on Monday and \(0.4\) on Sunday. If \(p\) drifts (fatigue, exam week), this check fails. All four hold → we write \(X \sim B(n, p)\).
\(X \sim B(n,\ p)\)All four checks pass for the diary week — if nights are independent, \(X \sim B(7,\ 0.4)\). Fail even one check and today's formula is off-limits. Let's test that "if":
Quick check · the reader's objection
Commit first · today's summit
Core concept 2 · C2
The poll's two moves are the whole formula: count the orders, weight one order. And here's the twist — the "35 of 128" from two classes ago was this formula in disguise, with \(p = 0.5\). With the honest \(p = 0.4\): \(P(X = 3) = \binom{7}{3}(0.4)^3(0.6)^4 = 35 \times 0.064 \times 0.1296 \approx 0.29\).
\(\binom{n}{k}\) counts how many ways \(k\) successes can be placed among \(n\) trials — the combinations from the counting class. For the poll: \(\binom{3}{2} = 3\) orders (SSO, SOS, OSS).
\(\binom{n}{k}\)Any one specific order with \(k\) successes has probability \(p^k q^{n-k}\) — the multiplication rule, with \(q = 1 - p\) written down explicitly first. Every order with the same \(k\) weighs exactly the same.
\(p^k q^{n-k}\)The lesson's most common error: reporting \(p^k q^{n-k}\) alone. That's one sequence, not the event. For the diary week the miss is huge: one order weighs \(0.0083\); all 35 together weigh \(0.29\).
\(P(X = k) = \binom{n}{k} p^k q^{n-k}\)Checked conditions, counted orders, weighted one order — that's the entire formula. Watch the sequences get counted one by one:
See it · C2
Predict first: for \(n = 3\) trials, how many of the \(2^3 = 8\) sequences have exactly 2 successes? Find them in the enumeration — then watch each one carry the same weight \(p^2 q\), and the formula assemble itself.
p = 0.40, p²(1−p) = 0.0960, P(X = 2) = 0.2880
Before dragging: will the 3 change when p changes?
Practice · exactly k · fresh numbers on demand
Core concept 3 · C3
The formula gives exactly \(k\) — one bar. Real questions ask for ranges: "at least 5 short nights", "more than 4". A range is a sum of bars, and the entire game is deciding which bars are in.
"At most \(k\)" is \(P(X \leq k)\): sum the bars from \(0\) up to and including \(k\). That running total is the CDF \(F(k)\) from last class — same object, binomial bars.
\(P(X \leq k) = F(k)\)"At least \(k\)" is everything except "at most \(k-1\)": \(P(X \geq k) = 1 - F(k-1)\). One subtraction can replace summing many bars — the complement rule doing honest work.
\(P(X \geq k) = 1 - F(k-1)\)"More than 4" excludes 4; "at least 4" includes it. The two differ by exactly one bar, \(P(X = 4)\) — and that one bar is where most range answers go wrong. Translate the phrase before touching arithmetic.
\(P(X > k) \neq P(X \geq k)\)Translate the phrase, mark the boundary, then choose the cheaper route — sum the bars you want, or subtract the ones you don't. See the boundary move first, then watch me run the whole decision:
See it · C3
Predict first: slide between "at least 3" and "more than 3" — which single bar switches sides? That one bar is the entire difference between the two phrases.
P(X ≤ 3) = 0.5941 for X ∼ B(8, 0.40)
I do · watch the whole decision
Watch me decide every step out loud — where the boundary sits, and whether direct sum or complement is cheaper.
"At least 5" means 5 is included: \(P(X \geq 5)\) — the bars at \(5, 6, 7\). I write the formal statement before any arithmetic; the phrase, not the numbers, decides the boundary. ("More than 4" lands on the same three bars.)
Direct sum needs the bars at \(5, 6, 7\) — three terms. The complement \(1 - P(X \leq 4)\) needs five terms. Three beats five: I sum directly. (For "at least 1" I'd flip: the complement is a single term.)
\(P(X{=}5) = \binom{7}{5}(0.4)^5(0.6)^2\) \(= 21 \times 0.01024 \times 0.36\) \(\approx 0.0774\).
\(P(X{=}6) = \binom{7}{6}(0.4)^6(0.6)^1\) \(= 7 \times 0.004096 \times 0.6\) \(\approx 0.0172\).
\(P(X{=}7) = \binom{7}{7}(0.4)^7(0.6)^0\) \(= (0.4)^7\) \(\approx 0.0016\).
\(P(X \geq 5) \approx 0.0774 + 0.0172 + 0.0016 = 0.096\). Between 0 and 1 ✓, small ✓ — a 5-plus-short-night week should be rare, and it is: about one week in ten.
The habit: translate the phrase, mark the boundary, count the terms each route needs, take the cheaper one. The boundary decision comes before any formula.
We do · you finish it
Same pipeline, one blank left for the room: you pick the route and supply the final probability.
Same model, \(X \sim B(7,\ 0.4)\). The phrase is "at least 1 short night this week." From the warm-up we already know \(P(X = 0) = (0.6)^7 \approx 0.028\).
Finish it: \(P(X \geq 1) = ?\)
Practice · ranges and boundaries · fresh numbers on demand
Now you run the whole thing solo — fresh numbers every re-roll, the \(F(x)\) row on screen, support at its minimum.
Core concept 4 · C4
Last class, \(E(X)\) meant a weighted sum over every value. For a binomial, the sum collapses: two tiny formulas give the same answers the long way would.
\(\mu = E(X) = np\) \(= 7 \times 0.4 = 2.8\) short nights per week. It's the same weighted-sum \(E(X)\) as last class — the binomial's structure just collapses the whole sum to one product. (The warm-up's "2.8" was this number, wearing a disguise.)
\(\mu = np\)\(\sigma_X = \sqrt{np(1-p)}\) \(= \sqrt{7(0.4)(0.6)}\) \(= \sqrt{1.68} \approx 1.30\). A typical week lands about \(1.3\) nights away from the average of \(2.8\) — so 1-to-4-short-night weeks are ordinary territory.
\(\sigma_X = \sqrt{np(1-p)}\)Two classic slips: \(\sqrt{np}\) forgets the \((1-p)\) inside the root, and \(np(1-p)\) unrooted is the variance, not the SD. The root goes around all three factors — or the wobble you report is wrong.
\(\sigma_X \neq \sqrt{np}\)Center \(np\), spread \(\sqrt{np(1-p)}\) — the two numbers every binomial hands over for free. Your turn to compute both:
Practice · mean & SD · fresh numbers on demand
Core concept 5 · C5
A binomial isn't one picture — it's a family. Where \(p\) sits relative to \(\tfrac{1}{2}\) decides which way the picture leans, and \(n\) decides how smooth it looks.
Successes are rarer than failures, so the mass piles up on low counts and a long tail stretches right. The diary model \(B(7, 0.4)\) leans gently right of its peak.
Success and failure trade places freely, so the picture is perfectly symmetric around \(np\) — the coin-flip world from the counting class, where all \(2^n\) sequences weigh the same.
Now failures are the rare thing: mass piles up on high counts, tail stretches left. Flip every success/failure label and the two skews are mirror images of each other.
And the second dial: hold \(p\) fixed and raise \(n\) — every binomial, however tilted, irons out toward a symmetric bell. Hold that thought; it's next class's door.
See it · C5
Predict first: start near \(n = 10,\ p = 0.3\) — which way does it lean? Now slide \(p\) to 0.7 and watch the lean flip. Then double \(n\): what happens to the tilt — at any \(p\)?
n = 10, p = 0.30, μ = 3.00, σ = 1.45, max P = 0.27 — shape: right-skewed
Core concept 6 · C6
Two outcomes and a count of successes do not make a binomial. Each BINS check has a signature way of failing — and most have a proper model waiting to take over.
Draw cards, people, or items from a small pool without putting them back and every draw changes the next one's odds. The right tool is the hypergeometric model. (Rule of thumb: binomial is a fair approximation only when the sample is under about 5% of the population.)
Flipping, calling, or testing until something happens has no \(n\) fixed in advance — the number of trials is itself the random outcome. That's the geometric model's territory, not the binomial's.
Quiz questions that get harder, players who tire, learners who improve — \(p\) changes trial to trial. The trap: averaging the \(p\)'s and pretending. No named shortcut applies; you're back to first-principles from last class.
The discipline that makes the formula trustworthy is the willingness to not use it. Test scenarios against the checklist, then try one that mixes an earlier class in:
See it · C6
Predict first: for each scenario, name the failing letter before revealing the verdict. Which failures point to a replacement model, and which leave you with no shortcut at all?
A 75% free-throw shooter keeps attempting shots until she makes 3. Count how many shots she takes.
Draw 5 cards from a standard 52-card deck without replacement. Count how many are hearts.
Flip a fair coin 10 times. Count the number of heads.
A machine's defect rate drifts upward during a shift. Inspect 20 items in sequence and count the defects.
Mark each BINS condition as passes or fails for this scenario, then press Check.
Which tool applies? · mixes an earlier class + today
Practice · binomial or not? · fresh numbers on demand
Production task · write the reply
Exit ticket · 60 seconds · anonymous
What Maya files
The "35 of 128" from the counting class was a binomial in disguise (\(p = 0.5\)). With the honest \(p = 0.4\): \(P(X = 3) = \binom{7}{3}(0.4)^3(0.6)^4 \approx 0.29\) — counting × weighting, and only after BINS grants the license.
Ranges are boundary decisions: "at least 5" sums three bars (\(\approx 0.096\) — one bad week in ten), "at least 1" flips to a one-term complement (\(1 - 0.028 = 0.972\)). And the week's summary comes free: \(\mu = np = 2.8\), \(\sigma = \sqrt{np(1-p)} \approx 1.30\).
The honest part is the fine print: the model assumes independent nights — the exact claim the reader disputes — and some settings (drawing from small pools, "until it happens", drifting \(p\)) get no binomial at all. Naming the assumption is what makes the number journalism, not numerology.
One thread left dangling: raise \(n\) and every binomial — tilted or not — smooths into the same bell. Next class, that bell gets a name: the normal distribution — and Maya finally checks whether her 300 sleep-hour responses actually follow it.
Vote counts on today's slides are simulated for rehearsal; live sessions show the room's real votes.