Binomial distribution — PMF, mean, variance: Foundations

This page assumes you have seen none of the notation. We build each symbol from the ground up, one picture at a time, so that when you re-read the parent note nothing is a mystery.

0. The scene: repeated yes/no trials

Picture a row of light bulbs. You flip each switch. Each bulb either turns on (we call that a success, ✓) or stays off (a failure, ✗). That's it — the whole topic is about rows of ✓ and ✗.

Figure s01 (below) draws exactly this: six labelled bulbs in a row, green ✓ boxes for successes and gray ✗ boxes for failures, so you can see what "one experiment = a row of $n$ trials" means. Every later figure reuses this green-✓ / gray-✗ colour code. This is the B of BINS — Binary — made visible: only two colours appear.

Figure — Binomial distribution — PMF, mean, variance

Why do we need this vocabulary? Because the parent note keeps saying "success/fail" — and if you don't picture that row of ✓/✗, none of the counting later makes sense.

1. The symbol $p$ — the chance of a single success

The picture: think of $p$ as the fraction of a bar painted the success colour. A bar that is $0.7$ blue means each trial has a $70%$ chance of ✓.

Why a number in $[0, 1]$ ? Because a probability is a share of certainty. Zero share = impossible; full share ( $1$ ) = guaranteed; anything in between is a partial chance. Nothing can be more certain than certain, so it never exceeds $1$ .

Because $p$ is the same on every trial, this is the S of BINS — Same probability. If $p$ drifted from trial to trial, the neat formulas would collapse.

The parent note writes $(1 - p)$ everywhere and sometimes shortens it to $q$ . They are the same thing — the un-painted part of the bar.

$p + (1 - p) = 1 ⟺ p + q = 1.$

2. The symbol $n$ — how many trials

The picture is literally the length of the row of bulbs. Ten bulbs → $n = 10$ .

Why must it be fixed in advance? Because if you kept flipping "until you got bored," the count would depend on your mood, not on chance — and the tidy formulas would break. Fixing $n$ is the N of BINS — Number fixed.

3. The symbol $X$ and $k$ — the count, and a value it can take

So " $X = k$ " is shorthand for the English sentence: "the count of successes came out equal to $k$ ." And " $P (X = k)$ " reads "the probability that we get exactly $k$ successes."

Figure s02 (below) shows a five-trial row with three ✓; a blue arrow says "count the checks," landing on the orange result $X = 3$ , i.e. the value $k = 3$ . Its pedagogical job: to separate the rule $X$ from the value $k$ visually, so the notation $P (X = k)$ stops looking abstract.

Why separate $X$ from $k$ ? $X$ is the rule ("count the ✓"); $k$ is a particular answer we plug in. Keeping them apart lets us write one PMF and reuse it for every possible answer $k = 0$ up to $k = n$ .

4. Independence — why we get to multiply

The picture and the rule: for independent events, the probability of "this AND that AND that" is the product of the separate probabilities.

$P (✓ then ✓ then ✓) = p \times p \times p = p^{3} .$

Why does this matter so much? The parent note's very first PMF step multiplies $p$ 's and $q$ 's together. That multiplication is only legal because trials are independent — this is the I of BINS, Independent. See Bernoulli distribution for the single-trial atom this is all built from.

5. Powers — $p^{k}$ and $q^{n - k}$ in plain words

So when the parent writes $p^{k}$ , it means: " $k$ trials each succeeded, so multiply $p$ by itself $k$ times." And $q^{n - k} = (1 - p)^{n - k}$ means: "the remaining $n - k$ trials each failed, so multiply $q$ that many times."

Why $n - k$ ? If $k$ of the $n$ trials succeeded, then $n - k$ of them (all the rest) failed. The two exponents always add up to $n$ : $k + (n - k) = n$ , one factor per trial.

$k successes p^{k} \times n - k failures (1 - p)^{n - k} = probability of ONE particular row of ✓ / \times .$

6. The factorial $n!$ — counting arrangements

The picture: $n!$ counts the number of ways to line up $n$ distinct objects in order. First slot: $n$ choices; next slot: $n - 1$ left; and so on down to $1$ .

Why do we need this? Because to build $(k n)$ next, we need a tool that counts arrangements — and $n!$ is exactly that tool.

7. The binomial coefficient $(k n)$ — "how many ways?"

This is the star symbol of the whole topic, so we build it slowly.

Figure s03 (below) draws all $6$ patterns of two ✓ among four boxes for $n = 4, k = 2$ — laid out as a grid so you can literally count them, and confirm the formula returns that same $6$ . Its job: to make $(k n)$ concrete as "the number of distinct ✓-placements" before any formula.

Look at the figure: with $n = 4$ slots and $k = 2$ successes, there are exactly $6$ distinct patterns of two ✓ among four boxes. We want a formula that spits out that $6$ without drawing every pattern.

WHY this formula (built from the pieces above):

$n!$ orders all $n$ slots — but that over-counts, because we don't care about the order among the $k$ chosen ones, nor among the $n - k$ leftover ones.
Divide by $k!$ to erase the ordering among the successes.
Divide by $(n - k)!$ to erase the ordering among the failures.

Check with the figure ( $n = 4, k = 2$ ): $(2 4) = \frac{4 !}{2 ! 2 !} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6. ✓$

Why does the topic need $(k n)$ ? Because many different rows give the same count $k$ , and each has the same probability $p^{k} q^{n - k}$ . So the true chance of "exactly $k$ successes" is (that probability) $\times$ (how many rows), and $(k n)$ supplies the second factor. This is exactly the parent note's Binomial Theorem ingredient.

8. The sum symbol $\sum$ — adding over all cases

The picture: line up the probability bars for $k = 0, 1, \dots, n$ and glue them end to end.

Why the topic needs it: to check the PMF is honest, you add the probabilities of every possible count. Because some count must happen, they must total exactly $1$ : $\sum_{k = 0}^{n} (k n) p^{k} (1 - p)^{n - k} = (p + (1 - p))^{n} = 1^{n} = 1.$

9. Expectation $E [X]$ — the long-run average, and why it is $n p$

The picture: it's the balance point of the probability bars — where the whole distribution would balance on a fingertip.

Now derive $E [X] = n p$ from zero. Break the whole count into one tiny counter per trial. Let $X_{i} = 1$ if trial $i$ succeeds and $X_{i} = 0$ if it fails. Then the total count is just the sum of these little counters: $X = X_{1} + X_{2} + \dots + X_{n} .$

Why this step? Adding a $1$ for each success is literally the same as counting successes.

The average of a single counter is easy — value times probability, added: $E [X_{i}] = success 1 \cdot p + failure 0 \cdot (1 - p) = p .$

Intuitive picture: on average each trial contributes a fraction $p$ of a success (a fair coin contributes $0.5$ per toss).

Now glue the $n$ trials together with linearity of expectation — no independence needed: $E [X] = E [X_{1}] + \dots + E [X_{n}] = n copies p + p + \dots + p = n p .$

$E [X] = n p$

So the mean is just "how much each trial pitches in ( $p$ ), times how many trials ( $n$ )."

10. Variance $Var (X)$ and $σ$ — how spread out, and why it is $n pq$

The picture: a narrow cluster of bars = small variance (outcomes bunch near the average); a wide spread = large variance.

Derive $Var (X) = n p (1 - p)$ step by step. Use the same little counters $X_{i}$ from section 9.

Step 1 — the wiggle of ONE counter. A handy shortcut: since $X_{i}$ is only ever $0$ or $1$ , squaring changes nothing ( $0^{2} = 0, 1^{2} = 1$ ), so $E [X_{i}^{2}] = E [X_{i}] = p$ . Then $Var (X_{i}) = E [X_{i}^{2}] - (E [X_{i}])^{2} = p - p^{2} = p (1 - p) = pq .$

Intuitive picture: $pq$ is biggest when $p = 0.5$ (a fair coin is the most unpredictable — maximum wiggle) and drops to $0$ when $p = 0$ or $p = 1$ (a certain trial has no wiggle at all).

Step 2 — add the wiggles. Here — unlike the mean — we need independence (the I of BINS). Only when trials don't interfere do their wiggles add without any cross-terms (covariances) creeping in; that is exactly the Variance and covariance rule. Adding $n$ identical wiggles: $Var (X) = Var (X_{1}) + \dots + Var (X_{n}) = n copies pq + \dots + pq = n pq = n p (1 - p) .$

$Var (X) = n p (1 - p), σ = n p (1 - p) .$

Why the topic needs this: knowing the average isn't enough — " $2$ defects expected" is far more useful when you also know "give or take $σ$ ."

Prerequisite map

The diagram below shows how the plain-word ingredients feed upward: the atomic trial splits into $p$ and $q$ ; $p$ , independence and powers build the per-row probability; $n!$ builds the coefficient; together they assemble the PMF, which then yields the mean and variance. Read it bottom-up as "what must I understand before this?"

Equipment checklist

Cover the right side and answer each before revealing.

What do the four letters BINS stand for?

Binary (two outcomes), Independent trials, Number

n

fixed, Same probability

p

What does

p

mean and what range can it take?

The probability of success on a single trial; a number in

[0, 1]

What is

q

in one line?

The failure probability

q = 1 - p

; together

p + q = 1

Why must

n

be fixed before starting?

So the count doesn't depend on when you stop; it's the "N" (Number fixed) in BINS.

Difference between

X

and

k

X

is the random count of successes;

k

is a specific value we ask about, as in

P (X = k)

What does "PMF" stand for and mean?

Probability mass function — the rule giving

P (X = k)

for every value

k

When are you allowed to multiply probabilities?

Only when events are independent — one outcome tells you nothing about another.

What does

p^{k}

mean in words?

Multiply

p

by itself

k

times — the chance that

k

chosen trials all succeed.

What is

0!

and

a^{0}

Both equal

1

(one way to arrange nothing; zero copies multiplied leaves

1

What does

(k n)

count, and its formula?

The number of ways to choose which

k

n

slots succeed;

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

Why does the PMF include

(k n)

Many equally-likely rows give

k

successes; the coefficient counts them all.

What is linearity of expectation and why care?

E [A + B] = E [A] + E [B]

for any quantities; it gives

E [X] = n p

without independence.

Derive

E [X] = n p

in one line.

X = \sum X_{i}

with

E [X_{i}] = p

, so

E [X] = n p

by linearity.

Derive

Var (X) = n p (1 - p)

in one line.

Each

X_{i}

has variance

p (1 - p)

; independence lets them add to

n p (1 - p)

For which

p

is the per-trial wiggle

pq

largest?

p = 0.5

; it is

0

p = 0

p = 1

Connections

Bernoulli distribution — the single trial ( $n = 1$ ) that every symbol here is built from.
Binomial Theorem — expands $(p + q)^{n}$ , supplying $(k n)$ and proving the PMF sums to $1$ .
Linearity of expectation — the tool that turns per-trial averages into $E [X] = n p$ .
Variance and covariance — why independence lets the per-trial wiggles add to $n p (1 - p)$ .
Poisson distribution — where the binomial heads when $n$ grows huge and $p$ shrinks.
Normal approximation to binomial — the smooth bell shape the bars approach for large $n$ .
Hypergeometric distribution — the without-replacement cousin (dependent trials break independence).