Binomial distribution — PMF, mean, variance: Visual walkthrough

Step 1 — One trial: the smallest possible experiment

WHAT. Before we flip $n$ coins, look at one. A single trial has exactly two doors it can walk through: success (we'll paint it teal) or failure (plum). We label the chance of success $p$ , a number between $0$ and $1$ . Since the two doors are the only two doors, the failure chance must be whatever is left over: $1 - p$ . We give that leftover its own name, $q = 1 - p$ , so we never have to write "one minus p" a hundred times.

WHY. A binomial experiment is nothing but this one trial, repeated. If we can't describe one flip honestly, everything downstream is built on sand. This single-trial object is the Bernoulli distribution.

PICTURE. Two boxes whose widths are the probabilities. Notice the widths must fill the whole strip — that is the statement $p + q = 1$ .

Figure — Binomial distribution — PMF, mean, variance

Step 2 — One specific sequence: multiply along the path

WHAT. Now flip the coin $n$ times. Write down one particular story — say the first $k$ flips are successes ( $S$ ) and the remaining $n - k$ are failures ( $F$ ):

k successes S S \dots S n - k failures F F \dots F

Because the flips don't influence each other, the probability of the whole story is the product of the probabilities of each step. Every $S$ contributes a factor $p$ ; every $F$ contributes a factor $q$ :

P (this exact story) = k copies of p p \cdot p \dots p \cdot n - k copies of q q \cdot q \dots q = p^{k} q^{n - k}

$p^{k}$ ::: $k$ successes, each multiplying in one factor of $p$ .
$q^{n - k}$ ::: the other $n - k$ trials failed, each multiplying in one factor of $q$ .

WHY multiply, and why are we allowed? Multiplication of probabilities is exactly what independence means: "the chance of A and B is the chance of A times the chance of B." That is the I in BINS. No independence → no multiplying → no clean formula.

PICTURE. A path down a tree, each edge labelled $p$ or $q$ ; the path's probability is the product of the edge labels it walks over.

Step 3 — Count the stories: where $(k n)$ comes from

WHAT. Every arrangement with $k$ successes — no matter where those successes sit — has the identical probability $p^{k} q^{n - k}$ (same number of $p$ 's, same number of $q$ 's, just shuffled). So $P (X = k)$ is that single probability multiplied by how many such arrangements exist. Counting them means: out of $n$ slots, choose which $k$ hold a success. That count is written $(k n)$ , read " $n$ choose $k$ ":

(k n) = \frac{n !}{k ! ( n - k )!}

$n!$ ::: all orderings of $n$ distinct slots.
$k!$ ::: divides out reorderings among the successes (we don't care which success is "first").
$(n - k)!$ ::: divides out reorderings among the failures for the same reason.

WHY this tool and not another? We asked a pure counting question — "in how many ways?" — with no order mattering. The mathematical object built precisely to answer "choose $k$ things from $n$ , order irrelevant" is the binomial coefficient. Using anything else (like $n^{k}$ , which does care about order) would over- or under-count.

PICTURE. For $n = 3, k = 2$ we lay out all the stories. Exactly $(2 3) = 3$ of them have two successes; each carries weight $p^{2} q^{1}$ .

Step 4 — The whole distribution, and the check that it sums to 1

WHAT. Sweep $k$ from $0$ to $n$ and you get a bar for each possible number of successes. Stack all the bars: their heights must add to exactly $1$ , because some number of successes definitely happens.

WHY it works out. The sum

k = 0 \sum n (k n) p^{k} q^{n - k}

is literally the expansion of $(p + q)^{n}$ by the Binomial Theorem — the theorem that says $(a + b)^{n} = \sum_{k} (k n) a^{k} b^{n - k}$ . Here $a = p$ , $b = q$ , and since $p + q = 1$ (Step 1!), we get

(p + q)^{n} = 1^{n} = 1. ✓

The two building blocks — the width picture $p + q = 1$ and the counting picture $(k n)$ — snap together perfectly.

PICTURE. The full bar chart for $n = 5, p = 0.5$ : a symmetric hump peaking near the middle, all bars summing to $1$ .

Step 5 — The mean, drawn as a balance point

WHAT. The mean $E [X]$ is the balance point of that bar chart — the spot where, if the bars were weights on a plank, it would tip level. We claim it sits at $n p$ .

WHY the slick proof beats brute force. Instead of the ugly sum $\sum_{k} k (k n) p^{k} q^{n - k}$ , split the total count into little pieces. Let $X_{i} = 1$ if trial $i$ succeeds, else $0$ — an indicator. Then

X = X_{1} + X_{2} + \dots + X_{n}

Each indicator's average is easy:

E [X_{i}] = 1 \cdot p + 0 \cdot q = p

$1 \cdot p$ ::: value $1$ (success) times its chance $p$ .
$0 \cdot q$ ::: value $0$ contributes nothing.

Now Linearity of expectation — the rule "the average of a sum is the sum of the averages," which holds even if the pieces interact — gives

E [X] = n terms p + p + \dots + p = n p

PICTURE. The bar chart with a triangular fulcrum planted at $x = n p$ ; the plank balances.

Step 6 — The variance, and why it needs independence

WHAT. Variance measures spread — how far the bars typically sit from the balance point. We claim it equals $n p (1 - p) = n pq$ .

WHY. First find the spread of one indicator. Because $X_{i}$ is only ever $0$ or $1$ , squaring changes nothing: $0^{2} = 0$ , $1^{2} = 1$ , so $X_{i}^{2} = X_{i}$ and $E [X_{i}^{2}] = p$ . Then

Var (X_{i}) = E [X_{i}^{2}] - (E [X_{i}])^{2} = p - p^{2} = p (1 - p) = pq

$E [X_{i}^{2}] = p$ ::: from the shortcut $X_{i}^{2} = X_{i}$ .
$(E [X_{i}])^{2} = p^{2}$ ::: the mean, squared.

Now adding the pieces' variances is only legal when the trials are independent — that's the I in BINS again, and it's what makes the covariance terms vanish. With independence:

Var (X) = n terms pq + pq + \dots + pq = n p (1 - p)

PICTURE. Two bar charts side by side: $p = 0.5$ (widest hump — maximum spread) versus $p = 0.9$ (skewed, narrow — small spread). The width of the hump is the variance.

Step 7 — The degenerate cases: the pictures still work

WHAT. Push $p$ and $k$ to the edges and confirm nothing breaks.

$p = 0$ (never succeeds). $P (X = 0) = (0 n) 0^{0} 1^{n} = 1$ ; every other bar is $0$ . Mean $= n \cdot 0 = 0$ , variance $= n \cdot 0 \cdot 1 = 0$ . One single bar of height $1$ at $k = 0$ — a certainty, no spread.
$p = 1$ (always succeeds). Mirror image: one bar of height $1$ at $k = n$ . Mean $= n$ , variance $= 0$ .
$k = 0$ in general. $(0 n) = 1$ so $P (X = 0) = q^{n}$ — all trials must fail, one story only. This is exactly the "at least one" complement engine: $P (X \geq 1) = 1 - q^{n}$ .
$n = 1$ . $(k 1)$ gives just two bars: $q$ at $k = 0$ , $p$ at $k = 1$ . Mean $p$ , variance $pq$ — we have circled back to the single Bernoulli distribution of Step 1.

WHY show these. A formula you trust only in the "nice middle" is a formula you don't trust. The edges are where mistakes hide — here they all reduce to a single spike (a sure thing) or back to one Bernoulli trial.

PICTURE. Three tiny charts: the $p = 0$ spike, the $p = 1$ spike, and the $n = 1$ two-bar Bernoulli.

The one-picture summary

Everything on this page in a single frame: one trial splits into $p$ and $q$ (Step 1) → multiply along a story to get $p^{k} q^{n - k}$ (Step 2) → count stories with $(k n)$ (Step 3) → the bars sum to $1$ via $(p + q)^{n}$ (Step 4) → they balance at $n p$ (Step 5) → they spread by $n pq$ (Step 6).

Recall Feynman retelling — say it to a friend

Picture flipping a slightly-bent coin $n$ times. One flip has two outcomes whose chances $p$ and $q$ fill a strip end to end — they add to one. A whole run of flips is a path through a tree; because flips don't talk to each other, you multiply the little chances along the path, giving $p^{k} q^{n - k}$ for a run with $k$ heads. But lots of different runs have the same number of heads — you count them with " $n$ choose $k$ ," which is just "in how many ways can I place $k$ heads in $n$ slots." Multiply the count by the per-run weight and you have the whole bar chart. Those bars must sum to one, and they do, because summing them is secretly expanding $(p + q)^{n} = 1^{n} = 1$ . Where does the chart balance? Each flip pulls the average up by $p$ on its own, so $n$ flips give $n p$ — that's the mean, and it doesn't care whether flips interact. How wide is the hump? Each flip wobbles by $p (1 - p)$ , and because they're independent you're allowed to just add those wobbles: $n p (1 - p)$ . Set $p$ to $0$ or $1$ and the wobble dies — a sure thing has one tall bar and no spread.

Connections

Bernoulli distribution — Step 1 and the $n = 1$ degenerate case are Bernoulli.
Binomial Theorem — the engine behind "the bars sum to 1" in Step 4.
Linearity of expectation — the shortcut that gives $n p$ in Step 5.
Variance and covariance — why independence lets the wobbles add in Step 6.
Poisson distribution — what the bar chart becomes when $n \to \infty, p \to 0, n p$ fixed.
Normal approximation to binomial — the smooth curve the hump approaches for large $n$ .
Hypergeometric distribution — the without-replacement cousin where Step 2's multiplying breaks.