Visual walkthrough — Common distributions (Bernoulli, Binomial, Poisson)

This page rebuilds the whole family — the parent note's central result — starting from a single coin flip, entirely in pictures. By the end you will see why

$P(X=k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

is nothing but a very large pile of coin flips squinted at from far away. We build every symbol before we use it.

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Step 1 — A single flip is just two bars

WHAT. Imagine an experiment with only two endings. Call the good ending success and write it as the number $1$; call the other ending failure and write it as $0$. A quantity that lands on $0$ or $1$ depending on chance is called a random variable, and we name it $X$.

We give success a chance we call $p$ — a number between $0$ and $1$. ($p=0$ means it never succeeds, $p=1$ means it always succeeds.) Since the two chances must fill up the whole "$100\%$", failure gets whatever is left: $1-p$.

WHY these two bars and nothing else. A picture of a probability is a bar whose height is the chance. Two outcomes → two bars. There is no third bar because there is no third ending. The two heights have to add to $1$, or we would be claiming probability leaks out somewhere.

PICTURE. The left bar (height $1-p$) is failure at $x=0$; the right bar (height $p$) is success at $x=1$.

$P(X=1):::p$ $P(X=0):::1-p$

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Step 2 — Line up $n$ flips: a path of successes and failures

WHAT. Now flip the same coin $n$ times, one after another, and demand the flips are independent — earlier flips tell you nothing about later ones. We are no longer asking "did it succeed?" but "how many of the $n$ flips succeeded?" Call that count $X$; it can be $0,1,2,\dots ,n$.

WHY count, not list. A specific story — say S, F, S, S, F — is one path through the flips. If flips are independent, the chance of a story is the product of its flip-chances. A story with $k$ successes and $n-k$ failures always multiplies to the same number:

$\underset{k\text{ successes}}{\underbrace{p\cdot p\cdots p}}\times \underset{n-k\text{ failures}}{\underbrace{(1-p)\cdots (1-p)}}=p^k(1-p{)}^{n-k}.$

PICTURE. Each row below is one possible story of length $n=4$. The green cells are successes, coral cells failures. Notice several different-looking rows share the same success count.

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Step 3 — Count the stories: where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ comes from

WHAT. Since every $k$-success story has the same probability, the total chance of "exactly $k$ successes" is that single-story probability multiplied by how many such stories exist. That count has a name: $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, read "$n$ choose $k$" — the number of ways to pick which $k$ of the $n$ positions are the green ones.

$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!},n!=n\cdot (n-1)\cdots 2\cdot 1.$

WHY a factorial. To fill $k$ green slots you have $n$ choices for the first, $n-1$ for the second, ... — that's the top $n!$ shrinking. But the greens are interchangeable, so we divide out the $k!$ orderings among the greens and the $(n-k)!$ orderings among the coral. What's left is the number of genuinely different pictures.

PICTURE. Pascal-style: the tree of choices collapses because reshuffling identical colours gives the same row.

$\text{Sanity: }{\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}=(p+(1-p){)}^n=1^n=1.$

Every outcome is covered, nothing leaks — exactly the guarantee we insisted on in Step 1.

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Step 4 — Mean and spread without any calculus

WHAT. The count $X$ is literally the sum of $n$ single flips: $X=X_1+X_2+\cdots +X_n$, where each $X_i$ is a Bernoulli $0/1$. We want the average count (the mean, written $E[X]$) and how tightly counts cluster around it (the variance, $\text{Var}(X)$).

WHY summing helps. Averages add up no matter what — even for dependent things — so the mean of a sum is the sum of the means. Each flip averages $p$, and there are $n$ of them:

$E[X]=\underset{n\text{ times}}{\underbrace{p+p+\cdots +p}}=np.$

Spread is fussier: variances add only because we demanded independence in Step 2. Each flip has variance $p(1-p)$ (biggest at $p=\frac{1}{2}$, when the coin is most undecided), so

$\text{Var}(X)=\underset{n\text{ times}}{\underbrace{p(1-p)+\cdots +p(1-p)}}=np(1-p).$

PICTURE. The bar chart of a Binomial: a hump centred at $np$, its width set by $\sqrt{np(1-p)}$.

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Step 5 — Squint: keep the average, shrink the chance

WHAT. Now the big move. We want to model rare events — server errors, disease cases — where there is no natural "$n$". So we let $n$ grow huge while $p$ shrinks tiny, tied together so that the average number of successes stays a fixed value we call $\lambda$ (lambda):

$np=\lambda ⟹p=\frac{\lambda }{n}.$

WHY this trade. A rare event = "many, many chances, each almost never happening, but a sensible average count." That is exactly "$n$ big, $p$ small, $np$ fixed." $\lambda$ inherits the meaning of $E[X]=np$: the expected number of events.

PICTURE. Same expected hump position, but the fine structure of individual flips melts into a smooth rare-event profile as $n$ climbs.

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Step 6 — The limit: three pieces, three destinations

WHAT. Substitute $p=\lambda /n$ into the Binomial and regroup into three chunks:

$P(X=k)=\underset{\text{(A) the counting head}}{\underbrace{\frac{n(n-1)\cdots (n-k+1)}{n^k}}}\cdot \underset{\text{(B) survives untouched}}{\underbrace{\frac{{\lambda }^k}{k!}}}\cdot \underset{\text{(C) becomes }{e}^{-\lambda }}{\underbrace{{\left(1-\frac{\lambda }{n}\right)}^n}}\cdot \underset{\text{(D)}\to 1}{\underbrace{{\left(1-\frac{\lambda }{n}\right)}^{-k}}}$

WHY each piece goes where it goes as $n\to \infty$ (with $k$ fixed and small):

• (A) $\frac{n(n-1)\cdots (n-k+1)}{n^k}$: a ratio of $k$ factors, each just below $1$ (like $\frac{n-1}{n}$). Each $\to 1$, so the product $\to 1$.
• (B) $\frac{{\lambda }^k}{k!}$: no $n$ in it — it just watches.
• (C) ${\left(1-\frac{\lambda }{n}\right)}^n\to e^{-\lambda }$. This is the definition of $e^{-\lambda }$: the number $e$ is what ${\left(1+\frac{x}{n}\right)}^n$ approaches, here with $x=-\lambda$. Why exponential enters at all: it is the unique thing a "constant tiny nibble taken $n$ times" converges to — compound-interest of vanishing failures.
• (D) ${\left(1-\frac{\lambda }{n}\right)}^{-k}$: base $\to 1$, fixed power $-k$, so $\to 1$.

PICTURE. Watch each of the four factors on its own axis as $n$ marches to the right.

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Step 7 — Edge and degenerate cases (never leave a gap)

WHAT. We check the corners so the reader never meets a scenario we skipped.

$k=0$ (nothing happens). $P(X=0)=\frac{{\lambda }^0{e}^{-\lambda }}{0!}=e^{-\lambda }$, since ${\lambda }^0=1$ and $0!=1$. As $\lambda \to 0$ (events truly never occur) this $\to 1$: a clean interval is certain. As $\lambda$ grows, a quiet interval becomes very unlikely.

$\lambda \to 0$. All probability piles onto $k=0$ — the distribution collapses to "always zero," the degenerate rare-event limit.

Large $\lambda$. The hump drifts right to sit at $\lambda$ and spreads with width $\sqrt{\lambda }$ (recall $\text{Var}=\lambda$), starting to look like a smooth bell.

Mean = Variance. For Poisson both equal $\lambda$ — inherited by squinting at Step 4: $E=np\to \lambda$ and $\text{Var}=np(1-p)\to \lambda \cdot 1=\lambda$ because $(1-p)\to 1$. This equality is a diagnostic: if real data has variance far above its mean, Poisson is the wrong model.

PICTURE. Three Poisson bar charts for $\lambda =0.5,3,20$, showing the collapse, the classic hump, and the near-bell.

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The one-picture summary

One flip becomes a stack of flips becomes a smooth rare-event curve — the same probability mass, re-viewed at three zoom levels. Bernoulli's two bars, Binomial's hump at $np$, Poisson's hump at $\lambda$, all one continuous idea.

Recall Feynman retelling — say it like a story

A coin has two endings: it lands on $1$ with chance $p$, on $0$ with chance $1-p$. Those are two bars that must add to a whole. Flip it $n$ times and don't care about order, only how many heads. Any exact story with $k$ heads costs $p^k(1-p{)}^{n-k}$, and there are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such stories, so multiply: that's the Binomial. Its average is $np$ and its wobble is $np(1-p)$ — averages always add, wobbles add only because the flips ignore each other. Now imagine the coin flipped a zillion times with each head almost impossible, but the average number of heads pinned at $\lambda$. Substitute $p=\lambda /n$ and the formula splits into four factors: two of them politely fade to $1$, one is ${\lambda }^k/k!$ that just sits there, and the last, $(1-\lambda /n{)}^n$, curls up into $e^{-\lambda }$ — that's the very definition of $e$ at work. What's left is ${\lambda }^k{e}^{-\lambda }/k!$: the Poisson law for rare events, where the mean and the variance are the same number $\lambda$, and a quiet interval has probability $e^{-\lambda }$.

Recall

What ties $n$ and $p$ together in the Poisson limit? ::: We hold $np=\lambda$ fixed while $n\to \infty$, $p\to 0$. Which factor becomes $e^{-\lambda }$ and why? ::: ${\left(1-\frac{\lambda }{n}\right)}^n$, because that is the defining limit of the exponential. Why do we divide the Binomial by $k!$ (and Poisson too)? ::: To undo the interchangeable orderings of the $k$ successes so we count each distinct outcome once. Poisson's signature check? ::: Mean equals variance, both $=\lambda$.