Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Exercises

Level 1 — Recognition

Goal: read a word problem and match it to exactly one of the five distributions, then write down its probability mass function (the rule that gives $P (X = k)$ ).

Problem 1.1

A single fair die is rolled once. Let $X = 1$ if it shows a six, else $X = 0$ . Name the distribution of $X$ and state $P (X = 1)$ and $P (X = 0)$ .

Recall Solution 1.1

WHAT it is: one trial, two outcomes (six / not six). That is the Bernoulli atom. Success probability $p = \frac{1}{6}$ , so $q = 1 - \frac{1}{6} = \frac{5}{6}$ . $P (X = 1) = \frac{1}{6}, P (X = 0) = \frac{5}{6} .$ WHY not Binomial? Binomial needs several fixed trials; here $n = 1$ , and Binomial $(1, p)$ is Bernoulli $(p)$ .

Problem 1.2

You keep buying lottery tickets, each independently winning with probability $p = 0.02$ , until you win your first prize. Let $X$ be the number of tickets bought. Name the distribution and give $P (X = k)$ .

Recall Solution 1.2

Signal words: "until the first success." That is the Geometric distribution (trials-counting convention). With $p = 0.02$ , $q = 0.98$ : $P (X = k) = q^{k - 1} p = (0.98)^{k - 1} (0.02), k = 1, 2, 3, \dots$ WHY $q^{k - 1}$ ? The first $k - 1$ tickets must all lose, and the $k$ -th wins — the winning position is fixed at the end, so there is no "choose" factor.

Problem 1.3

A page of text averages $λ = 0.5$ typos. Let $X$ be the number of typos on one page. Name the distribution and give $P (X = k)$ .

Recall Solution 1.3

Signal: many tiny opportunities (each character could be a typo), rare event, a rate per unit. That is Poisson. See Poisson process. $P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !} = \frac{( 0.5 ) ^{k} e ^{- 0.5}}{k !}, k = 0, 1, 2, \dots$

Level 2 — Application

Goal: plug numbers into one pmf and compute.

Problem 2.1

A quiz has 12 true/false questions; you guess each independently with $p = 0.5$ . What is $P (exactly 9 correct)$ ?

Recall Solution 2.1

WHAT: fixed number of trials $n = 12$ , count successes → Binomial $(12, 0.5)$ . $P (X = 9) = (9 12) (0.5)^{9} (0.5)^{3} = (9 12) (0.5)^{12} .$ Now $(9 12) = (3 12) = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = 220$ and $(0.5)^{12} = \frac{1}{4096}$ . $P (X = 9) = \frac{220}{4096} \approx 0.0537.$ WHY $(9 12)$ ? It counts which 9 of the 12 questions you got right; each such arrangement has the same probability $(0.5)^{12}$ .

Problem 2.2

Calls arrive at $λ = 4$ per minute. What is $P (exactly 2 calls in one minute)$ ?

Recall Solution 2.2

Poisson $(λ = 4)$ , window = 1 minute (matches the rate, so no rescaling): $P (X = 2) = \frac{4 ^{2} e ^{- 4}}{2 !} = \frac{16 e ^{- 4}}{2} = 8 e^{- 4} \approx 0.1465.$

Problem 2.3

A basketball player sinks each free throw with $p = 0.7$ . What is the probability her first miss happens on the 4th throw?

Recall Solution 2.3

Here a "success" for the Geometric is the event we wait for — a miss, probability $0.3$ . Let $S = 0.3$ (miss) and let the "non-event" be a make, probability $0.7$ . First miss on throw 4 = three makes then a miss: $P = (0.7)^{3} (0.3) = 0.343 \times 0.3 = 0.1029.$ WHY re-label? Geometric counts trials until the event of interest. Here that event is the miss, so its probability plays the role of $p$ in $q^{k - 1} p$ : $(0.7)^{3} (0.3)$ .

Level 3 — Analysis

Goal: use means, variances, and cumulative ("at least / more than") reasoning.

Problem 3.1

For Binomial $(n = 20, p = 0.3)$ , compute the mean and variance, and the standard deviation.

Recall Solution 3.1

From the decomposition $X = X_{1} + \dots + X_{n}$ of independent Bernoullis (Linearity of expectation for the mean, independence for adding variances): $E [X] = n p = 20 (0.3) = 6, Var (X) = n pq = 20 (0.3) (0.7) = 4.2.$ Standard deviation $= 4.2 \approx 2.049$ . See Expectation and Variance.

Problem 3.2

You roll a fair die until the first six. What is $P (X > 3)$ — that you need more than 3 rolls? Give the exact value.

Recall Solution 3.2

Geometric $(p = \frac{1}{6})$ . "More than 3 rolls" means the first 3 rolls all fail: $P (X > 3) = q^{3} = (\frac{5}{6})^{3} = \frac{125}{216} \approx 0.5787.$ WHY just $q^{3}$ ? $X > 3$ is exactly the event "no six in the first three rolls" — three independent failures, each probability $\frac{5}{6}$ . This is the tail formula $P (X > m) = q^{m}$ .

Problem 3.3

Using the die-until-six setup, verify the Memorylessness property numerically: show $P (X > 5 ∣ X > 2) = P (X > 3) .$

Recall Solution 3.3

By the definition of conditional probability and the tail formula $P (X > m) = q^{m}$ : $P (X > 5 ∣ X > 2) = \frac{P ( X > 5 and X > 2 )}{P ( X > 2 )} = \frac{P ( X > 5 )}{P ( X > 2 )} = \frac{q ^{5}}{q ^{2}} = q^{3} .$ And $P (X > 3) = q^{3}$ directly. Both equal $(\frac{5}{6})^{3} = \frac{125}{216}$ . ✓ WHY they match: the two failures already spent ( $X > 2$ ) cancel out of the ratio. The die "forgets" them — this is the discrete cousin of the Exponential distribution's memorylessness.

Level 4 — Synthesis

Goal: combine distributions, or switch between conventions and derivations.

Problem 4.1

A quality inspector tests components (each defective independently with $p = 0.15$ ) until she finds the 3rd defective. Let $X$ be the number of components tested. Find $P (X = 6)$ and $E [X]$ .

Recall Solution 4.1

"Until the 3rd success" → Negative Binomial $(r = 3, p = 0.15)$ , with success = defective. $P (X = 6) = (3 - 1 6 - 1) p^{3} q^{6 - 3} = (2 5) (0.15)^{3} (0.85)^{3} .$ Now $(2 5) = 10$ , $(0.15)^{3} = 0.003375$ , $(0.85)^{3} = 0.614125$ : $P (X = 6) = 10 \times 0.003375 \times 0.614125 \approx 0.02073.$ Mean: a Negative Binomial is the sum of $r$ independent Geometrics, so $E [X] = \frac{r}{p} = \frac{3}{0.15} = 20.$ WHY $(2 5)$ and not $(3 6)$ ? The 6th (last) test is forced to be the stopping defective. Only the other $r - 1 = 2$ defectives are free to sit anywhere among the first $5$ tests.

Problem 4.2

Derive the Geometric mean $E [X] = 1/ p$ from scratch using a Geometric series differentiation, and confirm it for the die ( $p = \frac{1}{6}$ ).

Recall Solution 4.2

WHAT we start with: $E [X] = \sum_{k = 1}^{\infty} k q^{k - 1} p$ . WHY differentiate: the sum $\sum_{k = 1}^{\infty} k x^{k - 1}$ is just the term-by-term derivative of $\sum_{k = 0}^{\infty} x^{k} = \frac{1}{1 - x}$ (valid for $∣ x ∣ < 1$ ). Differentiating gives us the extra factor of $k$ we need. $\frac{d}{d x} (\frac{1}{1 - x}) = \frac{1}{( 1 - x ) ^{2}} = \sum_{k = 1}^{\infty} k x^{k - 1} .$ Put $x = q$ (so $1 - x = p$ ): $E [X] = p \cdot \frac{1}{( 1 - q ) ^{2}} = p \cdot \frac{1}{p ^{2}} = \frac{1}{p} .$ For the die, $p = \frac{1}{6} \Rightarrow E [X] = 6$ rolls on average. ✓

Problem 4.3

A page has an average of $λ_{1} = 0.5$ typos. A book has 300 such pages. What is the expected number of typos in the book, and the probability the book has exactly 0 typos over its whole length? (Typos on different pages are independent.)

Recall Solution 4.3

Key idea: the sum of independent Poissons is again Poisson, with rates added. Over 300 pages the rate scales: $λ_{book} = 300 \times 0.5 = 150.$ Expected typos $= 150$ . Probability of exactly 0 typos: $P (X = 0) = \frac{λ ^{0} e ^{- λ}}{0 !} = e^{- 150} \approx 7.18 \times 1 0^{- 66} .$ WHY scale $λ$ : this is the same window-scaling from L2 — enlarge the window (1 page → 300 pages) and the mean count grows proportionally, a property of the underlying Poisson process.

Level 5 — Mastery

Goal: chain several tools, verify a signature identity, and interpret.

Problem 5.1

A factory's daily defect count is Poisson $(λ = 3)$ . On a given day, find $P (X \geq 2)$ (at least two defects). Then, treating each day independently, find the probability that the first day with $\geq 2$ defects is day 4.

Recall Solution 5.1

Step 1 — Poisson tail (WHAT & WHY complement): computing $P (X \geq 2)$ directly needs infinitely many terms, so subtract the two easy ones from 1: $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) = 1 - e^{- 3} - 3 e^{- 3} = 1 - 4 e^{- 3} \approx 0.8009.$ Step 2 — days become a Geometric: call a day a "success" if it has $\geq 2$ defects, so the per-day success probability is $s = 1 - 4 e^{- 3} \approx 0.8009$ . The first success on day 4 is Geometric: $P = (1 - s)^{3} s = (4 e^{- 3})^{3} (1 - 4 e^{- 3}) \approx (0.1991)^{3} (0.8009) \approx 0.006318.$ WHY two layers: the within-day count is Poisson; the across-days waiting is Geometric built on a single yes/no event per day — a clean example of nesting the five distributions.

Problem 5.2

Confirm the Poisson signature fact for $λ = 3$ numerically: compute the mean and variance from the definition truncated at $k = 0, \dots, 40$ and show both equal $3$ to high accuracy.

Recall Solution 5.2

WHAT: $E [X] = \sum_{k} k \frac{λ ^{k} e ^{- λ}}{k !}$ and $Var (X) = E [X^{2}] - (E [X])^{2}$ . The parent derivation gives $E [X] = λ = 3$ and $Var (X) = λ = 3$ exactly. Truncating the sums at $k = 40$ (the tail beyond that is astronomically small for $λ = 3$ ) reproduces $3.000000$ for both. Signature: mean $=$ variance is the fingerprint of Poisson; if a dataset's variance far exceeds its mean, Poisson is the wrong model (it is "over-dispersed"). This equality is inherited from the Binomial limit where $n pq = n p (1 - p) \to λ$ as $p \to 0$ .

Problem 5.3

Let $X \sim Binomial (n = 5, p = 0.4)$ . Use the MGF $M_{X} (t) = (q + p e^{t})^{n}$ to recover $E [X]$ , and check it equals $n p$ .

Recall Solution 5.3

WHY the MGF: derivatives of $M_{X} (t)$ at $t = 0$ generate moments — $M_{X}^{'} (0) = E [X]$ — turning an infinite sum into calculus. $M_{X} (t) = (q + p e^{t})^{n} \Rightarrow M_{X}^{'} (t) = n (q + p e^{t})^{n - 1} p e^{t} .$ At $t = 0$ , $e^{0} = 1$ and $q + p = 1$ : $M_{X}^{'} (0) = n \cdot 1^{n - 1} \cdot p = n p = 5 (0.4) = 2.$ This matches $E [X] = n p = 2$ . ✓

Recall One-line self-quiz

Geometric $P (X > m)$ shortcut? ::: $q^{m}$ — the first $m$ trials all failed. Negative Binomial "choose" factor? ::: $(r - 1 k - 1)$ — last trial is the forced stopping success. Poisson mean vs variance? ::: Both equal $λ$ . Scaling a Poisson to a bigger window? ::: Multiply $λ$ by the window ratio.