Common distributions (Bernoulli, Binomial, Poisson): Exercises

This is a self-testing page for Common distributions (Bernoulli, Binomial, Poisson). Try each problem with the Solution callout collapsed, then open it to check every step. The difficulty climbs in five levels:

Before we start, a short list of the notation, so no symbol appears unexplained.

Definition Symbols Used On This Page

$X$ = the random variable: the number that our experiment produces (a count of successes or events). It always takes whole-number values $0, 1, 2, \dots$ — you can never have $2.5$ successes.
The shorthand $X \sim Binomial (n, p)$ reads " $X$ is distributed as Binomial with parameters $n$ and $p$ " — it just names which rule assigns probabilities to each value of $X$ .
PMF = probability mass function: a rule $P (X = k)$ that gives the probability that $X$ lands on the exact whole number $k$ . ("Mass" because each allowed value carries a lump of probability.)
The support is the set of values $X$ is allowed to take: for Bernoulli $k \in {0, 1}$ ; for Binomial $(n, p)$ , $k \in {0, 1, \dots, n}$ ; for Poisson, $k \in {0, 1, 2, \dots}$ with no upper limit. Sums over $k$ run over exactly this set.
$p$ = probability of a single success (one trial coming out "yes"). A number between $0$ and $1$ .
$n$ = number of independent trials (a whole counting number).
$k$ = the exact count of successes we ask about (a value in the support).
$λ$ (lambda) = average number of events per interval, for rare-event counting.
$(k n) = \frac{n !}{k ! ( n - k )!}$ = "n choose k", the number of ways to pick which $k$ of the $n$ trials succeed.
$e \approx 2.71828$ = the constant that shows up when rare events pile up; $e^{- λ}$ shrinks as $λ$ grows.

Figure — Common distributions (Bernoulli, Binomial, Poisson)

Figure 1 — the decision map: use it whenever a problem asks "which distribution?" One single yes/no trial is Bernoulli; a fixed number $n$ of trials is Binomial; rare events with no fixed $n$ are Poisson.

Figure 2 — the shapes of the three PMFs used in these exercises. Notice Bernoulli is just two bars, Binomial $(10, 0.3)$ is a hump centred near $n p = 3$ , and Poisson $(4)$ has a long right tail — this skew is why Exercise 5.3 stays "common".

Figure 3 — the Binomial→Poisson limit from Exercise 4.1: as $n$ grows with $n p = 3$ fixed, the Binomial bars slide onto the Poisson $(3)$ curve. This is the picture behind "large $n$ , tiny $p$ ".

Level 1 — Recognition

Exercise 1.1 (L1)

A single user visits your landing page. They either sign up ( $1$ ) or leave ( $0$ ). The sign-up rate is $p = 0.2$ . Name the distribution, and give $P (sign up)$ and $P (leave)$ .

Recall Solution 1.1

What distribution? One trial, two outcomes → Bernoulli. We write $X \sim Bernoulli (0.2)$ , so the support is $X \in {0, 1}$ .

Why Bernoulli and not Binomial? Binomial needs many trials counted together; here there is exactly one user, one yes/no. That is the definition of a single Bernoulli trial.

$P (X = 1) = p = 0.2, P (X = 0) = 1 - p = 0.8.$

Read straight off the definition — nothing to compute.

Exercise 1.2 (L1)

Match each scenario to Bernoulli, Binomial, or Poisson: (a) Number of typos in a 300-page book. (b) Whether one packet is dropped by a router (drop / no-drop). (c) Number of heads in 20 coin flips.

Recall Solution 1.2

(a) Typos are rare events spread over a long "opportunity" (many characters), no fixed small $n$ — Poisson (support $k \in {0, 1, 2, \dots}$ , unbounded).
(b) A single yes/no event — Bernoulli (support ${0, 1}$ ).
(c) A fixed number of independent trials ( $n = 20$ ), counting successes — Binomial (support ${0, 1, \dots, 20}$ ).

Why (a) is Poisson, not Binomial: you could say each character is a Bernoulli trial with tiny error probability, but $n$ (number of characters) is huge and $p$ is tiny — that is exactly the limit where Binomial becomes Poisson.

Level 2 — Application

Exercise 2.1 (L2)

You send $n = 10$ emails, each clicked independently with $p = 0.3$ . Find the probability of exactly 3 clicks.

Recall Solution 2.1

Which distribution? Fixed $n = 10$ independent yes/no trials, count successes → Binomial. $X \sim Binomial (10, 0.3)$ , support $k \in {0, 1, \dots, 10}$ ; here $k = 3$ .

The PMF (probability mass function) and why each piece is there: $P (X = 3) = (3 10) (0.3)^{3} (0.7)^{7} .$

$(0.3)^{3}$ : three specific emails clicked (independence → multiply).
$(0.7)^{7}$ : the other seven not clicked.
$(3 10)$ : any 3 of the 10 could be the clicked ones, so count the arrangements.

Numbers: $(3 10) = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120,$ $(0.3)^{3} = 0.027, (0.7)^{7} \approx 0.0823543.$ $P (X = 3) = 120 \times 0.027 \times 0.0823543 \approx 0.2668.$

About 26.7%.

Exercise 2.2 (L2)

A server logs $λ = 4$ errors per hour on average. What is $P (exactly 2 errors)$ next hour?

Recall Solution 2.2

Which distribution? Events arrive continuously with a constant rate, no fixed $n$ → Poisson. $X \sim Poisson (4)$ , support $k \in {0, 1, 2, \dots}$ ; here $k = 2$ .

$P (X = 2) = \frac{λ ^{k} e ^{- λ}}{k !} = \frac{4 ^{2} e ^{- 4}}{2 !} = \frac{16 e ^{- 4}}{2} .$

With $e^{- 4} \approx 0.0183156$ : $P (X = 2) = 8 \times 0.0183156 \approx 0.1465.$

About 14.7%.

Exercise 2.3 (L2)

For $X \sim Binomial (50, 0.8)$ (a classifier tested on 50 samples at 80% accuracy), give the expected number correct and the standard deviation.

Recall Solution 2.3

Mean: $E [X] = n p = 50 \times 0.8 = 40$ . Variance: $Var (X) = n p (1 - p) = 50 \times 0.8 \times 0.2 = 8$ . Standard deviation: $σ = 8 \approx 2.83$ .

Interpretation: expect $40$ correct, with typical wobble of about $\pm 2.83$ around it.

Level 3 — Analysis

Exercise 3.1 (L3)

With $X \sim Poisson (3)$ server errors per hour, find $P (at least 1 error)$ .

Recall Solution 3.1

Why not sum $P (1) + P (2) + \dots$ ? The support runs $k = 1, 2, 3, \dots$ with no upper bound — that's an infinite sum. Instead use the complement: "at least 1" is the opposite of "exactly 0". $P (X \geq 1) = 1 - P (X = 0) .$ $P (X = 0) = \frac{3 ^{0} e ^{- 3}}{0 !} = e^{- 3} \approx 0.049787.$ $P (X \geq 1) = 1 - 0.049787 \approx 0.9502.$

About 95.0% — a clean hour is rare when you average 3 errors.

Exercise 3.2 (L3)

You flip a fair coin ( $p = 0.5$ ) 6 times. Find $P (at most 2 heads)$ .

Recall Solution 3.2

Why sum three terms? "At most 2" means $k = 0, 1, 2$ — these are mutually exclusive, so add their probabilities. $P (X \leq 2) = \sum_{k = 0}^{2} (k 6) (0.5)^{k} (0.5)^{6 - k} = (0.5)^{6} \sum_{k = 0}^{2} (k 6) .$ Every term shares the factor $(0.5)^{6} = 1/64$ because $p = 1 - p = 0.5$ . $(0 6) + (1 6) + (2 6) = 1 + 6 + 15 = 22.$ $P (X \leq 2) = \frac{22}{64} = 0.34375.$

About 34.4%.

Exercise 3.3 (L3)

A dataset of email counts has sample mean $\overset{x}{ˉ} = 5.0$ and sample variance $s^{2} = 12.0$ . Is Poisson a plausible model? Justify.

Recall Solution 3.3

The diagnostic: for Poisson, mean $=$ variance ( $E [X] = Var (X) = λ$ ). So compute the ratio $\frac{s ^{2}}{x ˉ} = \frac{12.0}{5.0} = 2.4.$ A Poisson fit predicts this ratio near $1$ . Here it is $2.4$ — the variance is much larger than the mean.

Verdict: Poisson is not a good model; the data is over-dispersed. A Negative Binomial (which allows variance $>$ mean) fits over-dispersed counts better.

Level 4 — Synthesis

Exercise 4.1 (L4)

$1000$ users each independently sign up with probability $p = 0.003$ . Approximate $P (exactly 2 sign-ups)$ two ways — exact Binomial vs Poisson — and compare (see Figure 3).

Recall Solution 4.1

Setup: $n = 1000$ , $p = 0.003$ , so $λ = n p = 3$ . Since $n$ is large and $p$ is tiny, Poisson should approximate Binomial well.

Poisson approximation ( $k = 2$ from the Poisson support ${0, 1, 2, \dots}$ ): $P (X = 2) \approx \frac{3 ^{2} e ^{- 3}}{2 !} = \frac{9 \times 0.049787}{2} \approx 0.2240.$

Exact Binomial ( $k = 2$ from the support ${0, 1, \dots, 1000}$ ): $P (X = 2) = (2 1000) (0.003)^{2} (0.997)^{998} .$ $(2 1000) = 499500$ , $(0.003)^{2} = 9 \times 1 0^{- 6}$ , $(0.997)^{998} \approx 0.049965$ . $P (X = 2) \approx 499500 \times 9 \times 1 0^{- 6} \times 0.049965 \approx 0.2246.$

Comparison: $0.2240$ vs $0.2246$ — agree to three decimals. This is the parent note's limit ( $n \to \infty, p \to 0, n p = λ$ ) working in practice, and it justifies using Poisson when Binomial is computationally awful.

Exercise 4.2 (L4)

Two independent web pages get errors at rates $λ_{A} = 2$ and $λ_{B} = 3$ per hour. Let $T = X_{A} + X_{B}$ be total errors. What distribution does $T$ follow, and find $P (T = 1)$ .

Recall Solution 4.2

Key fact (sum of independent Poissons): if $X_{A} \sim Poisson (2)$ and $X_{B} \sim Poisson (3)$ independently, then $T = X_{A} + X_{B} \sim Poisson (λ_{A} + λ_{B}) = Poisson (5),$ whose support is again ${0, 1, 2, \dots}$ . Why the rates add: errors from both pages are just "errors" arriving independently at a combined average rate $2 + 3 = 5$ per hour.

$P (T = 1) = \frac{5 ^{1} e ^{- 5}}{1 !} = 5 e^{- 5} \approx 5 \times 0.0067379 \approx 0.0337.$

About 3.4%.

Level 5 — Mastery

Exercise 5.1 (L5)

Prove that for $X \sim Bernoulli (p)$ , the variance equals $p (1 - p)$ , and show it is maximised at $p = 0.5$ .

Recall Solution 5.1

Step 1 — mean. $E [X] = 0 \cdot (1 - p) + 1 \cdot p = p$ .

Step 2 — $E [X^{2}]$ . Since $X$ takes only the support values $0$ or $1$ , $X^{2} = X$ , so $E [X^{2}] = E [X] = p$ .

Step 3 — variance. $Var (X) = E [X^{2}] - (E [X])^{2} = p - p^{2} = p (1 - p) .$

Step 4 — maximise. Let $f (p) = p - p^{2}$ . Differentiate and set to zero: $f^{'} (p) = 1 - 2 p = 0 \Rightarrow p = \frac{1}{2} .$ Since $f^{''} (p) = - 2 < 0$ , this is a maximum. The peak value is $f (\frac{1}{2}) = \frac{1}{4}$ .

Why $p = 0.5$ ? Variance measures uncertainty; it's largest when both outcomes are equally likely and smallest ( $0$ ) when $p = 0$ or $p = 1$ (a sure result).

Exercise 5.2 (L5)

Derive $E [X] = λ$ for $X \sim Poisson (λ)$ from the PMF, naming the tool at each step.

Recall Solution 5.2

Start from the definition of expectation (summed over the Poisson support $k = 0, 1, 2, \dots$ ): $E [X] = \sum_{k = 0}^{\infty} k \frac{λ ^{k} e ^{- λ}}{k !} .$

Step 1 — drop $k = 0$ . The $k = 0$ term is multiplied by $k = 0$ , so it vanishes: $E [X] = e^{- λ} \sum_{k = 1}^{\infty} k \frac{λ ^{k}}{k !} .$

Step 2 — cancel $k$ . Since $\frac{k}{k !} = \frac{1}{( k - 1 )!}$ : $= e^{- λ} \sum_{k = 1}^{\infty} \frac{λ ^{k}}{( k - 1 )!} .$

Step 3 — re-index with $j = k - 1$ (so $k = j + 1$ , $λ^{k} = λ \cdot λ^{j}$ ): $= e^{- λ} λ \sum_{j = 0}^{\infty} \frac{λ ^{j}}{j !} .$

Step 4 — collapse with the exponential series. The tool here is the Taylor series $e^{λ} = \sum_{j = 0}^{\infty} λ^{j} / j!$ : $= e^{- λ} λ e^{λ} = λ .$

Why this tool? The Poisson PMF is built from $e^{- λ}$ ; the exponential series is the one identity that makes an infinite sum of $λ^{j} / j!$ collapse to a closed form. No other tool does that cleanly. $■$

Exercise 5.3 (L5)

A disease affects $λ = 20$ people per year in a city. Using the complement, find $P (at least 25 cases)$ next year, i.e. $P (X \geq 25)$ , and state it to three decimals.

Recall Solution 5.3

Why the complement here? "At least 25" is an infinite tail over the support ${25, 26, \dots}$ ; the finite part $P (X \leq 24)$ is easier. $P (X \geq 25) = 1 - P (X \leq 24) = 1 - \sum_{k = 0}^{24} \frac{2 0 ^{k} e ^{- 20}}{k !} .$ Summing the $25$ terms (best done with a computer, as the numbers are unwieldy by hand) gives $P (X \leq 24) \approx 0.8432, P (X \geq 25) \approx 0.1568.$

About 15.7% — with mean $20$ and standard deviation $20 \approx 4.47$ , needing to exceed the mean by roughly one standard deviation lands near this modest probability.

Recall Quick self-check (reveal each answer)

Q: Poisson "at least one event" formula? A: $1 - e^{- λ}$

Q: When does Binomial collapse to Poisson? A: $n \to \infty, p \to 0, n p = λ$ fixed

Q: Variance of Bernoulli, and where is it maximised? A: $p (1 - p)$ , maximum at $p = 0.5$

Q: Sum of independent Poisson( $λ_{1}$ ) and Poisson( $λ_{2}$ )? A: Poisson( $λ_{1} + λ_{2}$ )