Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Worked examples

The scenario matrix

Every question in this topic lands in one of these cells. The worked examples below are labelled by the cell they hit, and together they cover all of them.

Cell	What makes it special	Which example
A. Binomial, ordinary	count successes, plain plug-in	Ex 1
B. Degenerate $p$	$p = 0$ or $p = 1$ — variance collapses	Ex 2
C. Boundary $k = 0$	"none happened" — often the easy leg of "at least one"	Ex 3
D. Geometric, ordinary	wait for first success + tail $P (X > k)$	Ex 4 (figure)
E. Geometric memorylessness	conditional "given you already waited"	Ex 5 (figure)
F. Negative Binomial	wait for $r$ -th success, the $(r - 1 k - 1)$ trap	Ex 6
G. Poisson, ordinary + window scaling	rate given per unit, asked over a different window	Ex 7
H. Limiting: Binomial $\to$ Poisson	large $n$ , small $p$ — show they agree	Ex 8
I. Word problem, mixed	choose the distribution yourself	Ex 9
J. Exam twist	mean/variance signature, or "at least" flips	Ex 10

Ex 1 — Cell A · Binomial, ordinary

Worked example Free throws

A player sinks each free throw independently with $p = 0.7$ . She takes $n = 5$ shots. What is $P (exactly 3 made)$ ?

Forecast: guess before reading — is it above or below $0.3$ ? (Her average is $n p = 3.5$ makes, so exactly $3$ should be fairly likely.)

Identify the distribution. Fixed number of trials $n = 5$ , count of successes → Binomial $(5, 0.7)$ . Why this step? "Fixed $n$ , count successes" is the signature of Binomial (parent §2).
Plug into $P (X = k) = (k n) p^{k} q^{n - k}$ with $k = 3$ , $q = 1 - 0.7 = 0.3$ : $P (X = 3) = (3 5) (0.7)^{3} (0.3)^{2} .$ Why this step? $(3 5) = 10$ chooses which 3 of the 5 shots go in; $(0.7)^{3}$ for those, $(0.3)^{2}$ for the two misses.
Compute: $(3 5) = 10$ , $(0.7)^{3} = 0.343$ , $(0.3)^{2} = 0.09$ , so $P = 10 \cdot 0.343 \cdot 0.09 = 0.3087.$

Verify (normalization, done by hand). A pmf must sum to $1$ , so let us actually add all six terms $k = 0..5$ with $p = 0.7, q = 0.3$ :

P (0) P (1) P (2) P (3) P (4) P (5) = 1 \cdot 0. 3^{5} = 0.00243, = 5 \cdot 0.7 \cdot 0. 3^{4} = 0.02835, = 10 \cdot 0. 7^{2} \cdot 0. 3^{3} = 0.13230, = 10 \cdot 0. 7^{3} \cdot 0. 3^{2} = 0.30870, = 5 \cdot 0. 7^{4} \cdot 0.3 = 0.36015, = 1 \cdot 0. 7^{5} = 0.16807.

Add them: $0.00243 + 0.02835 + 0.13230 + 0.30870 + 0.36015 + 0.16807 = 1.00000$ . ✓ Our $P (3) = 0.3087$ is exactly the middle term above, and it is indeed above the forecast threshold $0.3$ .

Verify (why the mode is $k = 4$ — the ratio test in full). Consecutive terms rise while the ratio exceeds $1$ : $\frac{P ( k )}{P ( k - 1 )} = \frac{( k n ) p ^{k} q ^{n - k}}{( k - 1 n ) p ^{k - 1} q ^{n - k + 1}} = \frac{n - k + 1}{k} \cdot \frac{p}{q} .$ Why this shape? Going from $k - 1$ to $k$ multiplies by one more $p$ , divides out one $q$ , and the choose-count changes by the factor $\frac{n - k + 1}{k}$ . Set this ratio $\geq 1$ and solve for $k$ : $(n - k + 1) p \geq k q \Rightarrow k \leq (n + 1) p$ . So terms grow up to $k = ⌊(n + 1) p ⌋ = ⌊ 6 \cdot 0.7 ⌋ = 4$ and shrink after — the mode is $4$ . From the table, $P (4) = 0.36015$ is indeed the largest, confirming $k = 3$ sits just left of the peak. ✓

Ex 2 — Cell B · Degenerate $p$

Worked example The certain and the impossible

Take Binomial $(4, p)$ . Evaluate the whole distribution when (i) $p = 1$ and (ii) $p = 0$ . What are the mean and variance in each case?

Forecast: if every trial always succeeds, how spread out can the count be?

Case $p = 1$ : $q = 0$ . Then $P (X = k) = (k 4) 1^{k} 0^{4 - k}$ . The factor $0^{4 - k}$ is zero unless $4 - k = 0$ , i.e. $k = 4$ . Why this step? $0$ raised to any positive power is $0$ , killing every term with $k < 4$ . The surviving term $k = 4$ has exponent $0^{0}$ , which we take to equal $1$ . Why is $0^{0} = 1$ here? In discrete-distribution formulas the factor $q^{n - k}$ is a count-of-failures bookkeeping term: when there are zero failures it should contribute nothing to the product, i.e. multiply by $1$ . Adopting the combinatorial convention $0^{0} = 1$ makes the single formula $(k n) p^{k} q^{n - k}$ stay correct at the boundaries $k = 0$ and $k = n$ without special-casing. (It is a convention, not a computed limit — chosen so the pmf sums to $1$ .) $P (X = 4) = 1, everything else 0.$
Mean & variance, $p = 1$ : $E [X] = n p = 4$ , $Var = n pq = 4 \cdot 1 \cdot 0 = 0$ . Why this step? Variance measures spread. A variable that is always $4$ has no spread — $Var = 0$ . (See Expectation and Variance.)
Case $p = 0$ : symmetric mirror. Now $q = 1$ and $p^{k}$ kills every term except $k = 0$ (using the same $0^{0} = 1$ for the $p^{0}$ factor). $P (X = 0) = 1$ , mean $= 0$ , variance $= n pq = 0$ . Why this step? Never succeeding means the count is certainly $0$ ; again a point mass, zero spread.

Verify: $Var = n pq = n p (1 - p)$ is a downward parabola in $p$ , zero at both ends $p = 0, 1$ and maximal at $p = \frac{1}{2}$ . Both degenerate cases correctly give variance $0$ . Units: probabilities dimensionless, sum to $1$ . ✓

Ex 3 — Cell C · Boundary $k = 0$ (the "at least one" trick)

Worked example At least one six

Roll a fair die $4$ times. What is $P (at least one six)$ ?

Forecast: each roll has a $1/6$ chance; over four rolls, is "at least one six" more or less than a half?

Complement first. "At least one" is annoying to sum ( $k = 1, 2, 3, 4$ ). Its complement is the single easy event "no sixes at all," i.e. $k = 0$ . Why this step? $P (\geq 1) = 1 - P (0)$ turns four terms into one. This is why the $k = 0$ boundary cell matters.
$P (0)$ from Binomial $(4, 1/6)$ : with $k = 0$ , $(0 4) = 1$ and $p^{0} = 1$ , so $P (X = 0) = (5/6)^{4} .$ Why this step? Zero successes means all four rolls fail; each fails with $q = 5/6$ , multiply (independence).
Subtract: $(5/6)^{4} = 625/1296 \approx 0.4823$ , so $P (\geq 1) = 1 - 0.4823 = 0.5177.$

Verify: Just over a half — matches the folk rule that ~ $4$ rolls gives you an even-money shot at a six. Cross-check: $1 - (5/6)^{4} = 671/1296$ . ✓

Ex 4 — Cell D · Geometric, ordinary + tail

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

Worked example Waiting for the first six

Roll a fair die until the first six appears. $X$ = the roll number it lands on. Find (a) $P (X = 3)$ , (b) $P (X > 3)$ , (c) $E [X]$ .

What the figure shows: a bar chart of $P (X = k)$ for $k = 1$ to $8$ . The bars shrink geometrically (each is $5/6$ of the one before). The two leftmost bars ( $k = 1, 2$ ) are drawn red — these are the forced failures before roll 3. The $k = 3$ bar is green — the target event $P (X = 3)$ . Everything from $k = 4$ rightward sits inside a pale shaded band labelled $P (X > 3) = q^{3}$ , i.e. the probability the six hasn't happened yet after three rolls.

Forecast: with $p = 1/6$ , roughly how many rolls until your first six?

Distribution. "Rolls until first success" → Geometric $(1/6)$ , so $q = 5/6$ (definition box above). Why this step? No "choose" appears: the success is forced to be last, so there's nothing to arrange.
(a) $P (X = 3) = q^{k - 1} p$ with $k = 3$ . In the figure, this is the green bar; the two red bars to its left are the required failures on rolls 1–2. $P (X = 3) = (5/6)^{2} (1/6) = 25/216 \approx 0.1157.$ Why this step? Two failures then one success — multiply the independent pieces along the timeline.
(b) Tail $P (X > 3)$ . $X > 3$ means the first three rolls are all failures (success hasn't happened yet). $P (X > 3) = q^{3} = (5/6)^{3} = 125/216 \approx 0.5787.$ Why this step? The shaded band in the figure = "still waiting after 3 rolls" = three failures in a row.
(c) Mean $E [X] = 1/ p = 6$ . Why this step? From the in-line intuition box above — chance $1/6$ per roll ⇒ wait $6$ rolls.

Verify: $P (X = 3) + P (X > 3)$ should equal $P (X \geq 3) = q^{2} = (5/6)^{2} = 25/36 \approx 0.6944$ ; indeed $0.1157 + 0.5787 = 0.6944$ . ✓ Mean $6$ matches the forecast.

Ex 5 — Cell E · Geometric memorylessness

Worked example Given you've already waited

Same die as Ex 4 ( $p = 1/6$ ). You've rolled $5$ times with no six yet. What is the probability you need more than $3$ additional rolls? Compare it to a fresh start.

What the figure shows: the top strip draws five red boxes marked "F" (your five past failures) with a dashed yellow line labelled "NOW" just after them. Below and to the right of NOW, a fresh set of blue bars — an identical Geometric distribution — sprouts as if the clock just reset. An arrow annotates that "P(more than 3 extra) $= q^{3} = P (X > 3)$ ": the five failures on the left simply drop out of the calculation.

Forecast: does the string of five failures make a six "due"? Trust your gut, then watch it be wrong.

Write the conditional. We want $P (X > 5 + 3 ∣ X > 5)$ . Why this step? "More than 3 additional" given "already past 5" is exactly the Memorylessness pattern $P (X > m + n ∣ X > m)$ with $m = 5, n = 3$ .
Definition of conditional probability: $P (X > 8 ∣ X > 5) = \frac{P ( X > 8 and X > 5 )}{P ( X > 5 )} = \frac{P ( X > 8 )}{P ( X > 5 )} .$ Why this step? ${X > 8}$ is already inside ${X > 5}$ , so the joint event is just ${X > 8}$ .
Use the tail $P (X > k) = q^{k}$ from the definition box in Ex 4: $= \frac{q ^{8}}{q ^{5}} = q^{3} = (5/6)^{3} \approx 0.5787.$ Why this step? The $q^{5}$ cancels — the five wasted rolls vanish. In the figure, sliding the origin to "NOW" gives an identical fresh geometric.
Compare: a brand-new player asking $P (X > 3)$ also gets $q^{3} \approx 0.5787$ (Ex 4b). Identical. Why this step? This is memorylessness: the die has no record of your suffering.

Verify: $q^{8} / q^{5} = q^{3}$ exactly, so the conditional equals the unconditional $P (X > 3)$ . The die is not "due." ✓

Ex 6 — Cell F · Negative Binomial (the $(r - 1 k - 1)$ trap)

Worked example Third success on the seventh trial

A drug trial: each patient responds independently with $p = 0.4$ . Let $X$ = the trial number on which the 3rd responder appears. Find $P (X = 7)$ and $E [X]$ .

Forecast: the average trial for the 3rd success — bigger or smaller than $7$ ?

Distribution. "Trials until the $r$ -th success" → Negative Binomial $(r, p)$ , $r = 3$ , $p = 0.4$ , $q = 0.6$ , with pmf $P (X = k) = (r - 1 k - 1) p^{r} q^{k - r}$ for $k \geq r$ . Why this step? And notice the edge case: at $r = 1$ this reduces to Geometric, since $(0 k - 1) = 1$ leaves $P (X = k) = p q^{k - 1}$ — exactly the Ex 4 pmf. Negative Binomial is "Geometric, but wait for the $r$ -th success."
The key insight — the last trial is forced. Trial $7$ must be a success (it's the 3rd, the stopping event). So among the first $6$ trials we place the other $2$ successes freely: $(2 6)$ , not $(3 7)$ . Why this step? This is mistake (a) from the parent. Fixing the last slot removes one degree of freedom.
Plug in, $k = 7$ : $P (X = 7) = (2 6) (0.4)^{3} (0.6)^{4} .$ Why this step? $p^{3}$ for the three responders, $q^{4}$ for the four non-responders, $(2 6) = 15$ arrangements of the two early successes.
Compute: $15 \cdot 0.064 \cdot 0.1296 = 0.124416$ .
Mean $E [X] = r / p = 3/0.4 = 7.5$ . Why this step? A Neg-Bin is $r$ independent Geometrics stacked; each waits $1/ p = 2.5$ trials, so $3 \times 2.5 = 7.5$ .

Verify: $E [X] = 7.5$ , so $P (X = 7)$ sitting just below the mean and being a solid peak-region value ( $\approx 0.124$ ) is consistent. Trap check: had we (wrongly) used $(3 7) = 35$ we'd get $0.29$ — obviously too big for a single point mass. ✓

Ex 7 — Cell G · Poisson with window scaling

Worked example Emails over two hours

Emails arrive as a Poisson process at rate $λ_{0} = 4$ per hour. What is $P (exactly 5 emails in the next 2 hours)$ ?

Forecast: over 2 hours you expect $8$ ; is exactly $5$ common or rare?

Scale $λ$ to the window. The rate is per hour, but the question spans $2$ hours, so $λ = λ_{0} \cdot t = 4 \cdot 2 = 8.$ Why this step? This is mistake (c) from the parent — the parameter is the expected count over the interval you're asked about, not the raw rate.
Plug into $P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !}$ , $k = 5$ , $λ = 8$ : $P (X = 5) = \frac{8 ^{5} e ^{- 8}}{5 !} = \frac{32768 e ^{- 8}}{120} .$ Why this step? Poisson pmf directly; $λ = 8$ is the mean over the 2-hour window.
Compute, one multiplication at a time. First $e^{- 8} \approx 3.3546 \times 1 0^{- 4}$ . Then the numerator: $32768 \times 3.3546 \times 1 0^{- 4} \approx 10.9924.$ Finally divide by $5! = 120$ : $P (X = 5) \approx \frac{10.9924}{120} \approx 0.0916.$

Verify: For Poisson mean $=$ variance $= 8$ , so standard deviation $\approx 2.83$ . The value $5$ is about one SD below the mean $8$ , so a probability near $0.09$ (below the peak $\approx 0.14$ at $k = 7, 8$ ) is right. Units: count is dimensionless. ✓

Ex 8 — Cell H · Limiting Binomial → Poisson

Worked example Rare defects, two ways

A factory makes $n = 1000$ chips; each is defective independently with $p = 0.002$ . Estimate $P (exactly 3 defective)$ two ways and show they agree.

Forecast: with expected defects $n p = 2$ , is exactly $3$ near the peak?

Exact Binomial $(1000, 0.002)$ : $P (X = 3) = (3 1000) (0.002)^{3} (0.998)^{997} \approx 0.18047.$ Why this step? This is the honest count model; but the binomial coefficient of a 4-digit $n$ is unwieldy by hand.
Poisson approximation. Since $n$ is large and $p$ is tiny, set $λ = n p = 1000 \cdot 0.002 = 2$ : $P (X = 3) \approx \frac{2 ^{3} e ^{- 2}}{3 !} = \frac{8 e ^{- 2}}{6} \approx 0.18045.$ Why this step? Parent §5 proved Binomial $(n, λ / n) \to$ Poisson $(λ)$ via $(1 - λ / n)^{n} \to e^{- λ}$ (see Limit e^x as (1+x/n)^n). Here $n = 1000$ is "large enough" that the two limits kicked in.
Compare: $0.18047$ vs $0.18045$ — agree to four decimals. Why this step? Confirms the limit isn't just theory: for rare events over many trials the messy binomial collapses to a one-parameter Poisson.

Verify: The relative error is under $0.02%$ . Both models have mean near $2$ ; Poisson variance $= 2$ , Binomial variance $= n pq = 2 \cdot 0.998 = 1.996$ — nearly equal because $p \to 0$ makes $q \to 1$ . ✓

Ex 9 — Cell I · Word problem (you pick the distribution)

Worked example Typos on a page

An editor finds typos at an average of $0.5$ per page. A chapter has $10$ pages. (a) What's the chance a given page has no typos? (b) What's the expected number of typos in the whole chapter, and its variance?

Forecast: most pages clean or most pages with typos?

Choose the model. Typos are rare events sprinkled over a continuum (the page) → Poisson, rate $λ_{page} = 0.5$ . Why this step? No fixed number of "trials," just a rate over an extent — that's the Poisson signature, not Binomial.
(a) $P (0)$ on one page, $k = 0$ , $λ = 0.5$ : $P (X = 0) = \frac{0. 5 ^{0} e ^{- 0.5}}{0 !} = e^{- 0.5} \approx 0.6065.$ Why this step? $k = 0$ boundary again — $λ^{0} = 1$ and $0! = 1$ leave just $e^{- λ}$ . So about $61%$ of pages are clean.
(b) Whole chapter. Scale $λ$ to $10$ pages: $λ_{chap} = 0.5 \cdot 10 = 5$ . Then $E [X] = 5, Var (X) = 5.$ Why this step? The chapter total is the sum of 10 independent per-page Poisson counts. A key structural fact of the Poisson family is that independent Poissons add: if $Y_{i} \sim$ Poisson $(λ_{i})$ independently, then $\sum Y_{i} \sim$ Poisson $(\sum λ_{i})$ . (One clean proof multiplies their probability generating functions $e^{λ_{i} (s - 1)}$ , whose product is $e^{(\sum λ_{i}) (s - 1)}$ — again a Poisson pgf.) So the chapter count is genuinely Poisson $(5)$ , not merely something with mean $5$ ; hence its variance is also $5$ (mean $=$ variance is the Poisson signature). Linearity of expectation alone would give the mean $5$ but could not tell us the whole distribution — we needed parameter-additivity for that.

Verify: About $61%$ of pages are clean — plausible for "half a typo per page." Since the chapter count is exactly Poisson $(5)$ , mean and variance both equal $5$ . ✓

Ex 10 — Cell J · Exam twist (mean = variance signature + "at least")

Worked example Which distribution has mean = variance?

A data set of counts has sample mean $\overset{x}{ˉ} = 3.0$ and sample variance $s^{2} = 3.0$ . An exam asks you to (a) name the natural model, then (b) using it, compute $P (X \geq 2)$ .

Forecast: does " $\geq 2$ " want a complement trick?

Read the signature. Mean $=$ variance is the fingerprint of Poisson (parent §5). So model as Poisson $(λ = 3)$ . Why this step? Binomial has variance $n pq < n p$ (since $q < 1$ ); Geometric has variance $q / p^{2} \neq = 1/ p$ in general. Only Poisson forces mean $=$ variance, and the data show exactly that ( $3.0 = 3.0$ ).
"At least 2" via complement. $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) .$ Why this step? Summing $k = 2, 3, 4, \dots$ is infinite; two subtractions is finite. This is the "at least" flip of the $k = 0$ boundary cell from Ex 3.
Compute the two easy terms, $λ = 3$ : $P (0) = \frac{3 ^{0} e ^{- 3}}{0 !} = e^{- 3}, P (1) = \frac{3 ^{1} e ^{- 3}}{1 !} = 3 e^{- 3} .$ Why this step? Direct Poisson pmf at $k = 0, 1$ ; $3^{0} = 1$ , $0! = 1! = 1$ .
Combine and factor $e^{- 3}$ : $P (X \geq 2) = 1 - e^{- 3} - 3 e^{- 3} = 1 - 4 e^{- 3} .$
Number: $e^{- 3} \approx 0.049787$ , so $4 e^{- 3} \approx 0.199148$ and $P (X \geq 2) = 1 - 0.199148 \approx 0.8009.$

Verify: Mean is $3$ , so seeing at least $2$ counts should be very likely — $\approx 0.80$ fits. Cross-check that probabilities balance: $P (0) + P (1) = 4 e^{- 3} \approx 0.1991$ , and $0.1991 + 0.8009 = 1.0000$ . ✓

Recall Rapid self-test (cover the answers)

Complement trick for "at least one six in 4 rolls" gives ::: $1 - (5/6)^{4} = 671/1296 \approx 0.5177$ Why does $(r - 1 k - 1)$ not $(r k)$ appear in Neg-Bin? ::: the last (k-th) trial is forced to be the stopping success, so only $r - 1$ successes are free among $k - 1$ trials Negative Binomial with $r = 1$ reduces to ::: the Geometric distribution Poisson rate $4$ /hr over $2$ hr uses $λ =$ ::: $8$ (scale rate by the window length) Memoryless conditional $P (X > 8 ∣ X > 5)$ for Geometric equals ::: $q^{3} = P (X > 3)$ — the past cancels Distribution with mean $=$ variance ::: Poisson Why is the sum of 10 independent Poissons still Poisson? ::: their parameters add (product of pgfs is again a Poisson pgf), not just their means