Binomial distribution — PMF, mean, variance: Worked examples

The scenario matrix

Every binomial question is one of these cells. Each worked example below is tagged with the cell it hits.

Cell	What it asks	Tool
A. Exact	$P (X = k)$ for one value of $k$	PMF once
B. At least one	$P (X \geq 1)$	complement $1 - (1 - p)^{n}$
C. At most / range	$P (X \leq m)$ or $P (a \leq X \leq b)$	sum a few PMF terms
D. Summary stats	mean, variance, standard deviation, "typical range"	$n p$ , $n p (1 - p)$
E. Extreme $p$	$p = 0$ or $p = 1$ (degenerate)	distribution collapses to a point
F. Extreme $k$	$k = 0$ or $k = n$ (the tails)	$(0 n) = (n n) = 1$
G. Not binomial	dependent trials / changing $p$	recognise & switch model
H. Poisson limit	huge $n$ , tiny $p$ , $n p$ moderate	$λ = n p$ approximation
I. Normal limit	large $n$ , middling $p$	$N (n p, n p (1 - p))$ + continuity

Figure — Binomial distribution — PMF, mean, variance

Every worked example below names the exact cell it lands on, so you can watch the tree get exercised end to end.

Example 1 — Cell A (exact $k$ )

Worked example Exactly 3 sixes in 6 rolls of a fair die

A fair die is rolled $n = 6$ times. "Success" $=$ rolling a six, so $p = \frac{1}{6}$ . Find $P (X = 3)$ .

Forecast: guess — is this more or less than $10%$ ? (Sixes are rare, so getting three of them feels unlikely.)

Check BINS. Six fixed rolls, each six/not-six, independent, same $p = \frac{1}{6}$ . Binomial. Why this step? Never apply the PMF before confirming the four conditions (Binary, Independent, Number fixed, Same $p$ ) — cell G exists precisely because people skip this.
Pick the PMF. We want one exact count $k = 3$ , so this is cell A: apply the PMF once. Why this step? Cell A asks for a single value of $k$ , and the PMF $(k n) p^{k} (1 - p)^{n - k}$ is exactly the tool that returns the probability of one specific count — no summing needed. $P (X = 3) = (3 6) (\frac{1}{6})^{3} (\frac{5}{6})^{3} .$
Count the orderings. $(3 6) = \frac{6 !}{3 ! 3 !} = 20$ . Why this step? Three sixes can sit in $20$ different positions among the six rolls; each ordering has the same probability, so we multiply.
Multiply the pieces. $P (X = 3) = 20 \cdot \frac{1}{216} \cdot \frac{125}{216} = \frac{2500}{46656} \approx 0.0536.$ Why this step? We combine the count of orderings ( $20$ ) with the shared probability of one ordering, $(\frac{1}{6})^{3} (\frac{5}{6})^{3} = \frac{125}{46656}$ , exactly as the PMF prescribes.

Verify: $\approx 5.4%$ — below $10%$ , matching the forecast. Units: a probability, dimensionless, and $0 \leq 0.0536 \leq 1$ . ✓

The full shape of this distribution is the histogram in Figure 2, with the $k = 3$ bar highlighted.

Example 2 — Cell B (at least one)

Worked example At least one bullseye

An archer hits the bullseye with probability $p = 0.2$ on each independent shot. She takes $n = 10$ shots. Find $P (at least one bullseye)$ .

Forecast: with 10 tries at 20% each, does "at least one" feel closer to $50%$ or to $90%$ ?

Recognise cell B. "At least one" means $X \geq 1$ , i.e. $k = 1, 2, \dots, 10$ — nine terms. Why this step? Summing nine PMF values is slow; the complement is one line (this is the 80/20 move from the parent note).
Flip to the complement. $P (X \geq 1) = 1 - P (X = 0) .$ Why this step? The only way to not get at least one is to get exactly zero — a single easy term.
Compute $P (X = 0)$ . $(0 10) = 1$ , so $P (X = 0) = (0.8)^{10} \approx 0.1074.$ Why this step? At $k = 0$ there is only one arrangement (all misses), so the coefficient is $1$ and only the product of ten failure probabilities remains.
Subtract. $P (X \geq 1) = 1 - 0.1074 = 0.8926$ . Why this step? Total probability is $1$ , so whatever is not "zero bullseyes" must be "at least one bullseye".

Verify: $\approx 0.893$ , comfortably near $90%$ — matches the forecast. Sanity: as $n$ grows, $(0.8)^{n} \to 0$ so this probability rises toward $1$ , exactly as intuition demands. ✓

Example 3 — Cell C (at most / range)

Worked example At most 2 defects

A production line yields defects with $p = 0.05$ . In a sample of $n = 8$ , find $P (X \leq 2)$ .

Forecast: defects are rare, so "at most 2" should be very likely — above $95%$ ?

Recognise cell C. "At most 2" $= k \in {0, 1, 2}$ — a short sum. Why this step? $P (X \leq m)$ is not a single PMF value; it is a cumulative probability, so we must add every allowed count. Here only three terms qualify, so direct addition is quicker than any complement.
Term $k = 0$ . $(0 8) (0.95)^{8} = (0.95)^{8} \approx 0.6634$ . Why this step? One arrangement (all good), so the coefficient is $1$ and every one of the 8 items must be non-defective.
Term $k = 1$ . $(1 8) (0.05) (0.95)^{7} = 8 \cdot 0.05 \cdot 0.6983 \approx 0.2793$ . Why this step? $(1 8) = 8$ — the single defect can be any of the 8 items, and each such arrangement carries one $0.05$ and seven $0.95$ factors.
Term $k = 2$ . $(2 8) (0.05)^{2} (0.95)^{6} = 28 \cdot 0.0025 \cdot 0.7351 \approx 0.0514$ . Why this step? $(2 8) = 28$ counts the pairs of positions the two defects can occupy, each with two $0.05$ and six $0.95$ factors.
Add. $0.6634 + 0.2793 + 0.0514 \approx 0.9941$ . Why this step? The three counts $k = 0, 1, 2$ are mutually exclusive outcomes, so the probability of "at most 2" is their sum.

Verify: $\approx 0.994$ — above $95%$ , matching the forecast. Cross-check: $P (X \geq 3) = 1 - 0.9941 = 0.0059$ , a small tail, exactly what "rare defects" predicts. ✓

Example 4 — Cell D (summary statistics)

Worked example Typical range of correct answers

A student guesses on $n = 40$ four-option multiple-choice questions, so $p = 0.25$ . Give the mean, variance, standard deviation, and a "typical" range of correct answers.

Forecast: on 40 questions at 25%, roughly how many correct on average? And is $20$ correct plausible by luck?

Mean (cell D). $E [X] = n p = 40 (0.25) = 10$ . Why this step? Linearity of expectation: each question contributes $p$ correct on average, and there are 40 of them.
Variance. $Var (X) = n p (1 - p) = 40 (0.25) (0.75) = 7.5$ . Why this step? Trials are independent, so variances add — each Bernoulli contributes $p (1 - p)$ .
Standard deviation. $σ = 7.5 \approx 2.739$ , using the definition above.
Typical range. $E [X] \pm 2 σ = 10 \pm 5.48$ , i.e. roughly $5$ to $15$ correct. Why this step? Most mass of a bell-ish distribution lies within two standard deviations of the mean.

Verify: $20$ correct is $(20 - 10) /2.739 \approx 3.65$ standard deviations above the mean — extremely unlikely by pure guessing, so a score of 20 signals real knowledge. Units: "correct answers", a count. Mean $10 < n = 40$ . ✓

The mean and the $\pm 2 σ$ band are drawn over the distribution in Figure 3.

Example 5 — Cell E (degenerate $p$ )

Worked example When randomness disappears

A biased coin has $p = 1$ (always heads). It is flipped $n = 7$ times. Find the distribution of $X =$ number of heads. Then repeat for $p = 0$ .

Forecast: if it always lands heads, how much uncertainty is left?

Plug $p = 1$ into the PMF. $P (X = k) = (k 7) (1)^{k} (0)^{7 - k}$ . Why this step? We test the formula at its boundary to see it stay sane.
Zero kills every term except $k = 7$ . For $k < 7$ the factor $(0)^{7 - k} = 0$ ; only $k = 7$ gives $(0)^{0} = 1$ (the convention above). So $P (X = 7) = 1$ , all other probabilities $0$ . Why this step? A factor of $0$ wipes out any term, so the whole probability piles onto the one outcome with no failure factor.
Check the summary stats. $E [X] = n p = 7$ , $Var = n p (1 - p) = 7 \cdot 1 \cdot 0 = 0$ . Why this step? Zero variance means no spread — the outcome is certain, matching a spike at $7$ .
Mirror case $p = 0$ . Now every term with $k > 0$ carries $(0)^{k} = 0$ ; only $k = 0$ survives, $P (X = 0) = 1$ , and $Var = n \cdot 0 \cdot 1 = 0$ . Why this step? By the same zero-kills-terms logic with success and failure swapped, all mass lands on "no heads".

Verify: both variances are $0$ — the "max at $p = 0.5$ , zero at the ends" rule from the parent note holds exactly. The distribution collapses to a single point, no randomness left, matching the forecast. ✓

Example 6 — Cell F (extreme $k$ , both tails)

Worked example The two tails of a coin experiment

Flip a fair coin $n = 5$ times ( $p = 0.5$ ). Find $P (X = 0)$ and $P (X = 5)$ , and confirm they are equal.

Forecast: for a fair coin, should "all tails" and "all heads" be equally likely?

Bottom tail $k = 0$ . $(0 5) = 1$ , so $P (X = 0) = (0.5)^{0} (0.5)^{5} = (0.5)^{5} = \frac{1}{32} \approx 0.03125$ . Why this step? $k = 0$ is the "all failures" corner — one arrangement, five failure factors.
Top tail $k = 5$ . $(5 5) = 1$ , so $P (X = 5) = (0.5)^{5} = \frac{1}{32} \approx 0.03125$ . Why this step? $k = n$ is the "all successes" corner — again one arrangement, five success factors.
Why equal here? Because $p = 1 - p = 0.5$ : swapping successes for failures leaves each term's value unchanged. Why this step? Symmetry only appears at $p = 0.5$ ; for biased coins the tails differ.

Verify: both equal $\frac{1}{32}$ . If instead $p = 0.3$ : $P (X = 0) = 0. 7^{5} \approx 0.1681$ but $P (X = 5) = 0. 3^{5} \approx 0.00243$ — no longer equal, confirming symmetry is special to $p = 0.5$ . ✓

Example 7 — Cell G (NOT binomial)

Worked example The without-replacement trap

A box has $10$ balls: $4$ red, $6$ blue. You draw $3$ balls without replacement. Find $P (exactly 2 red)$ . Can you use binomial?

Forecast: does $p$ (chance of red) stay the same on each draw?

Test BINS. After removing a red ball, only $3$ reds remain out of $9$ — so $p$ changes from $\frac{4}{10}$ to $\frac{3}{9}$ . Why this step? The "Same $p$ " and "Independent" conditions both fail; binomial is illegal here.
Switch to Hypergeometric distribution. Count favourable over total: $P (2 red) = \frac{( 2 4 ) ( 1 6 )}{( 3 10 )} = \frac{6 \cdot 6}{120} = \frac{36}{120} = 0.3.$ Why this step? Choose $2$ of $4$ reds and $1$ of $6$ blues, over all ways to choose $3$ of $10$ .
Compare with the wrong binomial answer. Naively with $p = 0.4$ : $(2 3) (0.4)^{2} (0.6) = 0.288$ — close but wrong. Why this step? Seeing the two numbers side by side shows the size of the error you make by forcing binomial onto a dependent draw.

Verify: true answer $0.3$ ; the binomial shortcut gives $0.288$ , an error of $0.012$ . Small $n$ relative to population size is why they're near but not equal. ✓

Example 8 — Cell H (Poisson limit)

Worked example Rare typos in a long document

A typist makes an error with probability $p = 0.002$ per word, over $n = 1500$ words (independent). Find $P (exactly 2 errors)$ using the Poisson distribution approximation.

Forecast: big $n$ , tiny $p$ — how many errors on average, and how likely is exactly 2?

Recognise cell H. $n = 1500$ is huge, $p = 0.002$ is tiny, but $λ = n p = 3$ is moderate. Why this step? Summing binomial terms with $n = 1500$ is painful; the Poisson limit replaces it with one clean formula.
Set $λ = n p = 1500 \cdot 0.002 = 3$ . This is the expected number of errors. Why this step? The Poisson approximation is governed entirely by its mean $λ$ , which must match the binomial mean $n p$ .
Apply the Poisson PMF $P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}$ : $P (X = 2) = e^{- 3} \frac{3 ^{2}}{2 !} = e^{- 3} \cdot \frac{9}{2} \approx 0.2240.$ Why this step? As $n \to \infty, p \to 0$ with $n p = λ$ fixed, the binomial converges to Poisson, so we may read the answer off the simpler formula.

Verify: the exact binomial gives $(2 1500) (0.002)^{2} (0.998)^{1498} \approx 0.2242$ — agreeing to three decimals, confirming the approximation. ✓

Example 9 — Cell I (Normal approximation)

Worked example Polling with continuity correction

In a poll of $n = 100$ voters, each supports a measure with $p = 0.5$ . Find $P (X \leq 45)$ using the Normal approximation to binomial.

Forecast: $45$ is below the mean of $50$ — should this be well under $50%$ ?

Check the rule of thumb. The continuity-corrected Normal is only trustworthy when both $n p \geq 10$ and $n (1 - p) \geq 10$ . Here $n p = 50$ and $n (1 - p) = 50$ , both comfortably above $10$ . Why this step? Those conditions guarantee the distribution is tall and symmetric enough for a smooth bell to fit its bars; skipping the check is how people misuse the approximation on skewed, small- $n$ cases.
Recognise cell I. $n = 100$ is large and $p = 0.5$ is central, so the distribution is close to a bell curve — summing $46$ PMF terms by hand is absurd, and the Normal curve gives it in one lookup. Why this step? Choosing the Normal limit here is what turns a 46-term sum into a single standardised lookup.
Match mean and variance. $μ = n p = 50$ , $σ = n p (1 - p) = 25 = 5$ . Why this step? The approximating bell must share the binomial's centre and spread, or it would sit over the wrong place.
Continuity correction. Discrete $X \leq 45$ becomes continuous $X \leq 45.5$ . Why this step? A whole bar of width 1 sits over each integer; include the half-bar up to $45.5$ so the area matches.
Standardise using $Z$ . Let $Z$ be the standard normal random variable, $Z \sim N (0, 1)$ — a bell curve centred at $0$ with standard deviation $1$ . We convert our count into a $Z$ -value by subtracting the mean and dividing by $σ$ : $z = \frac{45.5 - 50}{5} = - 0.9.$ Why this step? Standardising lets us read any Normal probability off one universal $Z$ -table instead of a fresh curve for every $μ, σ$ .
Read the probability. From the standard-normal table, $P (Z \leq - 0.9) \approx 0.1841$ . Why this step? Once we have a single $z$ -value, the tail area to its left is the answer we wanted.

Verify: the exact binomial sum $P (X \leq 45) \approx 0.1841$ too — the continuity-corrected Normal nails it. And $0.184 < 0.5$ , matching the forecast that $45$ sits below the mean. ✓

The half-bar shift and the bell overlay are shown in Figure 4.

Active recall

Recall Which cell is "at least one", and why the complement?

Cell B. "At least one" spans many $k$ values, but its complement is the single easy term $P (X = 0) = (1 - p)^{n}$ , so $P (X \geq 1) = 1 - (1 - p)^{n}$ .

Recall What makes Example 7 not binomial, and what replaces it?

Drawing without replacement changes $p$ each draw (and breaks independence), so the "Same $p$ " condition fails. Use the Hypergeometric distribution instead.

Recall When do you reach for Poisson vs Normal?

Poisson when $n$ is huge and $p$ tiny with $λ = n p$ moderate (rare events). Normal when $n$ is large and $p$ is not extreme so that $n p \geq 10$ and $n (1 - p) \geq 10$ (bell-shaped) — remember the continuity correction and that $Z \sim N (0, 1)$ .

Connections

Binomial distribution — PMF, mean, variance — the parent formulas every example above applies.
Bernoulli distribution — each single trial in these examples.
Binomial Theorem — why the PMF sums to 1.
Linearity of expectation — the mean in Example 4.
Variance and covariance — why variances add in Example 4.
Poisson distribution — the rare-event limit in Example 8.
Normal approximation to binomial — the large- $n$ limit in Example 9.
Hypergeometric distribution — the correct model in Example 7.