Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Visual walkthrough

Before Step 1, let us agree on the four characters in this story. If any feels unfamiliar, that is fine — each gets a picture.

Step 1 — Chop the minute into $n$ tiny slots

WHAT. We take one continuous minute (a call could arrive at any instant) and pretend it is really $n$ separate, equal little time-slots. In each slot, at most one call happens: either a call (success, chance $p$ ) or silence (failure, chance $q = 1 - p$ ). That is exactly a Bernoulli trial per slot.

WHY. A Bernoulli flip is the only atom we know how to compute with. Time is continuous and scary; a row of $n$ independent flips is friendly. So we approximate the minute by $n$ flips and promise to fix the error later by taking $n \to \infty$ .

PICTURE. The red band is our single minute. Below it, the same minute cut into $n = 12$ slots — each is one coin flip.

Figure — Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial

If exactly $k$ of these slots hold a call, we have $k$ successes among $n$ flips — that is the Binomial from the parent note.

Step 2 — Write down the Binomial probability

WHAT. The chance of exactly $k$ successes in $n$ independent flips is the Binomial formula: $P (X = k) = which slots (k n) k calls p^{k} n - k silences (1 - p)^{n - k} .$

Term by term, right where it sits:

$(k n)$ — counts which of the $n$ slots hold the $k$ calls (order does not matter, so we "choose").
$p^{k}$ — each of the $k$ successful slots costs a factor $p$ ; multiplied because slots are independent.
$(1 - p)^{n - k}$ — each of the remaining $n - k$ silent slots costs a factor $q = 1 - p$ .

WHY. This is the exact answer for finite $n$ . Our job for the rest of the page is to see what it converges to.

PICTURE. One specific arrangement: 3 red (call) slots, 9 black (silent) slots. There are $(3 12)$ such arrangements.

Step 3 — Force the average to stay fixed: substitute $p = λ / n$

WHAT. We now impose the one rule that makes the limit meaningful: the average number of calls per minute, $n p$ , must equal a fixed number $λ$ . Solving $n p = λ$ gives $p = λ / n$ . Substitute it: $P (X = k) = (k n) (\frac{λ}{n})^{k} (1 - \frac{λ}{n})^{n - k} .$

WHY. If we just let $p \to 0$ without pinning $n p$ , all calls vanish and the answer is boringly $P (X = 0) = 1$ . If we let $p$ stay fixed while $n \to \infty$ , calls explode to infinity. Holding $n p = λ$ is the Goldilocks path: many slots, each with a tiny call-chance, but a steady average.

PICTURE. As we double $n$ (finer slots), each red slot gets shorter (smaller $p$ ), but the total red length — the expected number of calls $n p$ — stays exactly the same $λ$ .

Step 4 — Split the "choose" term open

WHAT. Expand $(k n) = \frac{n !}{k ! ( n - k )!}$ and pull the $(\frac{λ}{n})^{k} = \frac{λ ^{k}}{n ^{k}}$ apart. Regroup into three tidy chunks: $P (X = k) = \frac{λ ^{k}}{k !} \cdot chunk A \frac{n !}{( n - k )! n ^{k}} \cdot chunk B (1 - \frac{λ}{n})^{n} \cdot chunk C (1 - \frac{λ}{n})^{- k} .$

Where each piece came from:

$\frac{λ ^{k}}{k !}$ — the $λ^{k}$ is from $p^{k}$ ; the $\frac{1}{k !}$ is the leftover of $(k n)$ after we peel off the $n$ -parts. This chunk has no $n$ in it — it survives the limit untouched.
Chunk $A = \frac{n !}{( n - k )! n ^{k}}$ — the " $n$ -flavoured" leftover of the binomial coefficient.
Chunk $B = (1 - \frac{λ}{n})^{n}$ — the bulk of the silence term.
Chunk $C = (1 - \frac{λ}{n})^{- k}$ — the small correction because the exponent was $n - k$ , not $n$ .

WHY. We separate the algebra so that each chunk has a clean, known limit. Divide and conquer: solve each chunk in its own step.

PICTURE. The three chunks as three dials, each about to be turned as $n \to \infty$ . Only the red dial ( $B$ ) moves to a non-trivial place.

Step 5 — Chunk $A \to 1$ : the falling factorial

WHAT. Look closely at chunk $A$ : $\frac{n !}{( n - k )! n ^{k}} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - k + 1 )}{n ^{k}} = \frac{n}{n} \cdot \frac{n - 1}{n} \dots \frac{n - k + 1}{n} .$ There are exactly $k$ fractions, and as $n \to \infty$ each one $\to 1$ (the $- 1, - 2, \dots$ become negligible against a huge $n$ ). A product of $k$ ones is $1$ : $\frac{n !}{( n - k )! n ^{k}} ⟶ 1.$

Term by term:

$n! / (n - k)!$ — this is the product of the top $k$ integers $n, n - 1, \dots, n - k + 1$ (all the others cancel).
$n^{k}$ — one factor of $n$ for each of those $k$ terms, sitting underneath.
Each ratio $\frac{n - j}{n} = 1 - \frac{j}{n} \to 1$ because $\frac{j}{n} \to 0$ for fixed $j$ .

WHY. $k$ is a fixed small number (like 5), while $n$ races to infinity. Subtracting a fixed amount from an enormous number changes nothing proportionally.

PICTURE. The $k$ ratios plotted as $n$ grows — all of them climb toward the red horizontal line at height $1$ .

Step 6 — Chunk $B \to e^{- λ}$ : the exponential appears

WHAT. Chunk $B = (1 - \frac{λ}{n})^{n}$ is the star of the show. Its limit is the definition of the exponential: $(1 - \frac{λ}{n})^{n} ⟶ e^{- λ} .$ This is the standard limit $\left(1+\frac{x}{n}\right)^n \to e^{x}$ with $x = - λ$ .

WHY this tool and not another? We are multiplying a number slightly less than 1 by itself $n$ times, while $n \to \infty$ . Two runaway effects fight: the base $\to 1$ (which would give $1$ ) and the exponent $\to \infty$ (which would give $0$ ). Neither wins outright — the tension between them lands on a precise finite number, and that number is defined to be $e^{- λ}$ . No other elementary function captures "shrinking base to the power of exploding exponent"; the exponential exists precisely to answer this question.

PICTURE. The value of $(1 - \frac{λ}{n})^{n}$ (red curve) plotted against $n$ for $λ = 3$ , flattening onto the dashed line $e^{- 3} \approx 0.0498$ .

Step 7 — Chunk $C \to 1$ , and assemble the answer

WHAT. Chunk $C = (1 - \frac{λ}{n})^{- k}$ . Inside, $\frac{λ}{n} \to 0$ , so the base $\to 1$ ; raised to a fixed power $- k$ , it stays $\to 1^{- k} = 1$ . Now multiply the surviving pieces from Steps 4–6: $P (X = k) = \frac{λ ^{k}}{k !} \cdot A 1 \cdot B e^{- λ} \cdot C 1 = \frac{λ ^{k} e ^{- λ}}{k !} .$

Reading the final formula term by term:

$λ^{k}$ — expected count raised to the number we asked about; more calls asked $\Rightarrow$ higher power.
$e^{- λ}$ — the "nothing happened in the empty slots" factor, the fingerprint of the limit.
$\frac{1}{k !}$ — divides out the fact that the $k$ calls are indistinguishable in time (the survivor of $(k n)$ ).

WHY it's legitimate. A finite product of limits equals the limit of the product (each chunk converges), so we may take limits chunk-by-chunk and multiply.

PICTURE. The finished Poisson bars for $λ = 3$ : the whole probability mass, one red bar per value of $k$ .

Step 8 — Edge and degenerate cases (never leave a gap)

WHAT & WHY & PICTURE, three cases:

The one-picture summary

The whole derivation in a single flow: chop time → Binomial → pin $n p = λ$ → three chunks → $1 \times e^{- λ} \times 1$ → Poisson. The one non-trivial limit (red) is the exponential.

Recall Feynman retelling — the whole walkthrough in plain words

Picture one minute where a phone might ring. I cannot handle "any instant," so I chop the minute into a huge number $n$ of tiny slots and say each slot is a coin flip — ring or silence. The chance of a ring per slot is $p$ , and I insist the average number of rings, $n p$ , equals a fixed $λ$ (say 3). As I chop finer and finer, each slot's ring-chance $p = λ / n$ shrinks, but the total expected rings never changes. The exact Binomial answer splits into three pieces. Two of them are lazy — they just drift to $1$ as the slots multiply. The third piece, "all the silent slots multiplied together," is the interesting one: a number just under $1$ multiplied by itself infinitely often, landing exactly on $e^{- λ}$ . Multiply the survivors — $\frac{λ ^{k}}{k !}$ , times $1$ , times $e^{- λ}$ , times $1$ — and out drops $P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !}$ . Check it at $k = 0$ : just $e^{- λ}$ , the chance nothing rings. Add all $k$ up: the exponential series rebuilds $e^{λ}$ , cancels, gives $1$ . Perfect. That's Poisson: infinitely many microscopic coin flips, remembered by a single number $λ$ .

Recall Quick self-test

Why must we hold $n p = λ$ fixed instead of just letting $p \to 0$ ? ::: Otherwise all rings vanish and $P (X = 0) \to 1$ ; fixing $n p = λ$ keeps a steady average so the limit is meaningful. Which of the three chunks gives the exponential, and what is its limit? ::: Chunk $B = (1 - λ / n)^{n} \to e^{- λ}$ . What is $P (X = 0)$ for a Poisson? ::: $e^{- λ}$ . Why can chunk $A = \frac{n !}{( n - k )! n ^{k}}$ be sent to 1? ::: It is a product of $k$ ratios $\frac{n - j}{n} \to 1$ for fixed $j$ , and $k$ is fixed.