Code
set.seed(2026)6 Discrete Probability Distributions
Status: ported 2026-05-19. Reviewed by editor: pending.
Learning outcomes
By the end of this chapter the reader should be able to:
- Recognise the recurring patterns that motivate named discrete distributions and explain why a closed-form PMF is more useful than enumeration.
- Write the PMF, mean and variance of the Bernoulli, Binomial, Hypergeometric, Poisson and Geometric distributions.
- Identify which distribution applies in a real-world scenario (with or without replacement, fixed \(n\) or open-ended counting, etc.).
- Compute point and cumulative probabilities by hand and verify them in R using
dbinom,pbinom,dpois,ppois,dhyper,phyper,dgeom,pgeom. - Apply the Poisson approximation to the binomial and state the conditions under which it is accurate.
- Interpret the memoryless property of the geometric distribution and explain when it is a useful idealisation and when it is a modelling limitation.
Motivating empirical question
Given a chocolate factory with a 5% defect rate, what is the chance that an inspector who samples 20 bars finds more than three defectives — and how does the answer change if the inspector instead waits to see the first defective, or counts complaints per shift?
These three questions illustrate the three counting frameworks at the heart of this chapter: counting successes in a fixed number of trials (binomial), counting trials until the first success (geometric), and counting events in a fixed interval (Poisson). In each case the underlying experiment is different, but the recipe is the same: identify the random mechanism, match it to a named distribution, then read off the formulas for mean, variance and tail probabilities.
The chapter follows the chocolate-factory running example through every section, then pivots to insurance, lotteries, server crashes and startup pitching in the worked examples and exercises.
6.1 5.1 Why study named distributions?
In the previous topic we described a discrete random variable by listing every possible value together with its probability. That approach is general but exhausting: every new problem seems to require building a probability model from scratch. In practice a small number of patterns recur across economics, finance, quality control, epidemiology and marketing. Statisticians have given these recurring patterns names, and each named distribution comes with:
- A closed-form PMF \(p(x) = P(X = x)\) that lets you compute probabilities with a single formula instead of enumerating outcomes.
- Ready-made formulas for the mean \(\mu\) and variance \(\sigma^2\).
- Software functions in every modern toolkit (
dbinom,dpois,dhyper,dgeomin R;BINOM.DISTin Excel;scipy.statsin Python). - A body of theoretical results — limit theorems, approximations and relationships between distributions — that simplify analysis.
NoteRecognising a named distribution
Identifying that a real situation follows a known distribution is one of the most valuable skills in applied statistics. Once you make that identification, you gain immediate access to all the formulas, tables and software tools associated with that distribution.
The distributions we study in this chapter are listed below. The discrete uniform appears only briefly; we treat the rest in full subsections.
| Distribution | Setup in one line |
|---|---|
| Discrete Uniform | All \(k\) outcomes equally likely. |
| Bernoulli\((p)\) | A single yes/no trial. |
| Binomial\((n, p)\) | Number of successes in \(n\) independent trials. |
| Hypergeometric\((N, K, n)\) | Successes when sampling without replacement. |
| Poisson\((\lambda)\) | Events occurring at a constant average rate. |
| Geometric\((p)\) | Number of failures before the first success. |
The negative binomial distribution (number of failures before the \(r\)-th success) generalises the geometric; we mention it briefly at the end.
6.1.1 5.1.1 The discrete uniform (warm-up)
A random variable \(X\) follows a discrete uniform distribution on \(\{1, 2, \ldots, k\}\) if each value is equally likely:
\[ p(x) = P(X = x) = \frac{1}{k}, \qquad x \in \{1, 2, \ldots, k\}. \]
We write \(X \sim \mathrm{DU}(k)\). Its mean and variance are
\[ \mu = \frac{k+1}{2}, \qquad \sigma^2 = \frac{k^2 - 1}{12}. \]
The canonical example is a fair \(k\)-sided die: for \(k = 6\), \(\mu = 3.5\) and \(\sigma^2 = 35/12 \approx 2.917\). We will not dwell on it; it is included only because it is the simplest possible discrete model.
6.2 5.2 The Bernoulli distribution
Many situations involve a single trial with exactly two possible outcomes: a customer buys or does not buy, a product is defective or not, a loan defaults or does not. The Bernoulli distribution models this basic building block.
NoteDefinition: Bernoulli distribution
A random variable \(X\) follows a Bernoulli distribution with parameter \(p \in (0, 1)\) if it takes only two values: \(X = 1\) (“success”) with probability \(p\) and \(X = 0\) (“failure”) with probability \(1 - p\).
The PMF can be written compactly as
\[ p(x) = p^x (1 - p)^{1 - x}, \qquad x \in \{0, 1\}. \]
We write \(X \sim \mathrm{Bernoulli}(p)\). The mean and variance are
\[ \mu = \mathbb{E}[X] = p, \qquad \sigma^2 = \operatorname{Var}(X) = p(1 - p). \]
The variance \(p(1 - p)\) is maximised at \(p = 0.5\) (maximum uncertainty) and equals zero at \(p = 0\) or \(p = 1\) (certain outcome). The more balanced the two outcomes, the greater the variability.
NoteThree Bernoulli scenarios
- Coin toss. \(X = 1\) if heads, \(X = 0\) if tails. For a fair coin, \(p = 0.5\).
- Customer purchase. A visitor to an online store makes a purchase with probability \(p = 0.04\) (a 4% conversion rate). Then \(\mathbb{E}[X] = 0.04\) and \(\operatorname{Var}(X) = 0.04 \times 0.96 = 0.0384\).
- Defective item. A factory produces light bulbs with a 2% defect rate. Then \(p = 0.02\), \(\mathbb{E}[X] = 0.02\) and \(\operatorname{Var}(X) = 0.0196\).
6.3 5.3 The Binomial distribution
Now suppose we repeat a Bernoulli experiment \(n\) times and count the total number of successes. If every trial is independent and the probability of success \(p\) remains the same on each trial, the count follows a binomial distribution.
NoteDefinition: Binomial distribution
A random variable \(X\) follows a binomial distribution with parameters \(n\) (number of trials) and \(p\) (success probability) if its PMF is given by the formula below for \(k = 0, 1, \ldots, n\). We write \(X \sim B(n, p)\).
\[ p(k) = P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \qquad k = 0, 1, \ldots, n, \]
where \(\binom{n}{k} = \dfrac{n!}{k!\,(n - k)!}\) is the binomial coefficient. It counts the number of ways to choose which \(k\) of the \(n\) trials are successes; the term \(p^k (1 - p)^{n - k}\) gives the probability of any one such sequence.
The mean and variance follow directly from writing \(X = X_1 + \cdots + X_n\) as a sum of \(n\) independent Bernoullis:
\[ \mu = np, \qquad \sigma^2 = np(1 - p). \]
6.3.1 5.3.1 When does the binomial apply?
The binomial model applies whenever all three of the following conditions hold:
- Fixed number of trials \(n\), determined in advance.
- Constant probability \(p\) on every trial.
- Independence: the outcome of one trial does not affect the others.
WarningPitfall: sampling without replacement
If sampling is done without replacement from a finite population, the trials are not independent and \(p\) changes from trial to trial. The correct model is then the hypergeometric (Section 5.4). A common rule of thumb: if the sample is less than 5% of the population (\(n < 0.05 N\)), the binomial is an excellent approximation.
6.3.2 5.3.2 Shape of the binomial PMF
The shape of \(B(n, p)\) depends critically on \(p\). With \(n = 10\):
- \(B(10, 0.3)\) is right-skewed (most mass on small \(k\)).
- \(B(10, 0.5)\) is perfectly symmetric.
- \(B(10, 0.8)\) is left-skewed.
In general \(B(n, p)\) and \(B(n, 1 - p)\) are mirror images of each other.
6.3.3 5.3.3 Worked example: factory defects
A factory produces electronic components with a defect rate of 5%. An inspector randomly selects 20 components; let \(X\) be the number of defectives. Items are sampled from a large lot, so \(X \sim B(20, 0.05)\).
(a) \(P(X = 2)\). \[ P(X = 2) = \binom{20}{2}(0.05)^2 (0.95)^{18} = 190 \times 0.0025 \times 0.3972 = 0.1887. \]
(b) \(P(X \leq 1)\). \[ P(X = 0) = (0.95)^{20} = 0.3585, \quad P(X = 1) = 20 \times 0.05 \times (0.95)^{19} = 0.3774, \] so \(P(X \leq 1) = 0.7359\). About a 74% chance of at most one defective.
(c) Mean and standard deviation. \[ \mu = np = 1, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{0.95} \approx 0.9747. \]
6.3.4 5.3.4 Business example: online returns
An online retailer ships 50 orders per day; each is returned with probability \(p = 0.03\). Let \(X\) be the daily number of returns, so \(X \sim B(50, 0.03)\).
- Expected returns: \(\mathbb{E}[X] = 1.5\).
- Probability of zero returns: \(P(X = 0) = (0.97)^{50} = 0.2181\).
- Probability of more than three returns: \(P(X > 3) = 1 - P(X \leq 3) \approx 1 - 0.9372 = 0.0628\).
So even though the average is 1.5 returns, there is a 6.3% chance that the daily returns desk will be overwhelmed if it can only handle three.
6.4 5.4 The Hypergeometric distribution
The binomial assumes either sampling with replacement or sampling from a population so large that removing one item barely changes the probabilities. In many settings — inspecting a small lot, auditing a finite set of accounts, drawing cards — we sample without replacement. The hypergeometric handles this case.
NoteDefinition: Hypergeometric distribution
A finite population of size \(N\) contains \(K\) “successes” and \(N - K\) “failures”. A sample of \(n\) items is drawn without replacement. Let \(X\) be the number of successes in the sample. We write \(X \sim \mathrm{Hyp}(N, K, n)\).
\[ p(k) = P(X = k) = \frac{\displaystyle \binom{K}{k}\binom{N - K}{n - k}}{\displaystyle \binom{N}{n}}, \qquad \max(0, n - N + K) \leq k \leq \min(n, K). \]
The mean and variance are
\[ \mu = n \cdot \frac{K}{N}, \qquad \sigma^2 = n \cdot \frac{K}{N} \cdot \frac{N - K}{N} \cdot \frac{N - n}{N - 1}. \]
The mean has the same form as the binomial mean \(np\) with \(p = K/N\). The variance includes the extra factor \(\frac{N - n}{N - 1}\), called the finite population correction (FPC), which is always less than 1 (for \(n > 1\)). Sampling without replacement reduces variability: as you draw items, you acquire information about the population.
NoteWhen does hypergeometric reduce to binomial?
As \(N \to \infty\) with \(K/N \to p\) fixed and \(n\) fixed, the hypergeometric converges to \(B(n, p)\). Practical rule: if \(n < 0.05 N\), use the binomial as an approximation.
6.4.1 5.4.1 Worked example: lot inspection
A shipment of \(N = 20\) items contains \(K = 4\) defectives. An inspector draws \(n = 5\) items without replacement. The probability of finding exactly \(k = 2\) defectives is
\[ P(X = 2) = \frac{\binom{4}{2}\binom{16}{3}}{\binom{20}{5}} = \frac{6 \times 560}{15504} = 0.2167. \]
The expected number of defectives is \(\mu = 5 \times \frac{4}{20} = 1\), and the variance is \(5 \times 0.2 \times 0.8 \times \frac{15}{19} = 0.6316\).
6.4.2 5.4.2 Lottery example: Loter'ia Primitiva
In the Spanish Loter'ia Primitiva, a player picks 6 numbers from 1 to 49 and the draw produces 6 winning numbers. Let \(X\) be the number of matches between the player’s selection and the winning numbers, so \(X \sim \mathrm{Hyp}(49, 6, 6)\).
- \(P(X = 3) = \dfrac{\binom{6}{3}\binom{43}{3}}{\binom{49}{6}} = \dfrac{20 \times 12341}{13983816} = 0.01765\). About 1 in 57 tickets.
- \(\mathbb{E}[X] = 6 \times \frac{6}{49} = 0.7347\) matches.
- The jackpot: \(P(X = 6) = 1 / \binom{49}{6} \approx 7.15 \times 10^{-8}\), roughly 1 in 14 million.
6.5 5.5 The Poisson distribution
The Poisson distribution models the number of events that occur in a fixed interval of time, space, or other continuum, under the assumption that events occur independently and at a constant average rate.
NoteDefinition: Poisson distribution
A random variable \(X\) follows a Poisson distribution with parameter \(\lambda > 0\) if its PMF is the formula below for \(k = 0, 1, 2, \ldots\). We write \(X \sim \mathrm{Poisson}(\lambda)\).
\[ p(k) = P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \qquad k = 0, 1, 2, \ldots \]
The parameter \(\lambda\) is simultaneously the mean and the variance — an elegant and distinctive property:
\[ \mu = \lambda, \qquad \sigma^2 = \lambda. \]
This equidispersion property is the Poisson’s signature. Empirical count data that show variance much larger than the mean (“overdispersion”) signal that the Poisson is the wrong model — a common situation in applied work.
6.5.1 5.5.1 Conditions and typical scenarios
The Poisson applies when:
- Events occur independently of one another.
- The average rate is constant over the interval.
- Two events cannot occur at exactly the same instant (no clumping).
- The probability of an event in a tiny sub-interval is proportional to its length.
Typical applications include customers arriving at a bank per hour, typos per page, accidents per month at an intersection, emails per day, server crashes per year, and earthquakes of magnitude \(\geq 6.0\) per year in a region.
6.5.2 5.5.2 Worked example: call centre
A call centre receives on average \(\lambda = 3\) calls per minute. Let \(X\) be the number of calls in a randomly selected minute, so \(X \sim \mathrm{Poisson}(3)\).
- \(P(X = 5) = \dfrac{e^{-3} \cdot 3^5}{5!} = \dfrac{0.04979 \times 243}{120} = 0.1008\).
- \(P(X = 0) = e^{-3} = 0.04979\).
- \(P(X \leq 2) = e^{-3}\left(1 + 3 + \frac{9}{2}\right) = 0.04979 \times 8.5 = 0.4232\).
Scaling property. If \(X\) counts calls per minute with rate \(\lambda = 3\), then the number of calls in 2 minutes follows \(\mathrm{Poisson}(2 \times 3) = \mathrm{Poisson}(6)\). The rate is proportional to the length of the interval — one of the most useful features of the Poisson.
6.5.3 5.5.3 Shape of the Poisson PMF
For small \(\lambda\) the Poisson is heavily right-skewed (most mass at 0 and 1); as \(\lambda\) grows the distribution becomes more symmetric and bell-shaped. For \(\lambda \geq 10\) it closely resembles a normal distribution — a manifestation of the Central Limit Theorem we will explore in a later course.
6.6 5.6 The Poisson approximation to the Binomial
When the number of trials \(n\) is large and the success probability \(p\) is small, \(B(n, p)\) is well approximated by \(\mathrm{Poisson}(\lambda = np)\). This is both theoretically important and practically useful: computing \(\binom{n}{k}\) for large \(n\) can be tedious by hand.
NoteRule of thumb
If \(n \geq 30\) and \(p \leq 0.10\) (equivalently \(np \leq 10\)), then \(B(n, p) \approx \mathrm{Poisson}(np)\). The approximation improves as \(n\) increases and \(p\) decreases.
6.6.1 5.6.1 Worked example: website complaints
A large website receives 1000 visits per day. Each visit generates a complaint with probability \(p = 0.002\). Exact distribution: \(X \sim B(1000, 0.002)\). Approximation: \(\mathrm{Poisson}(\lambda = 2)\).
| \(k\) | \(B(1000, 0.002)\) | \(\mathrm{Poisson}(2)\) |
|---|---|---|
| 0 | 0.1353 | 0.1353 |
| 1 | 0.2707 | 0.2707 |
| 2 | 0.2709 | 0.2707 |
| 3 | 0.1806 | 0.1804 |
| 4 | 0.0902 | 0.0902 |
| 5 | 0.0361 | 0.0361 |
The two columns are virtually identical. The expected number of daily complaints is \(\lambda = 2\), with \(\sigma = \sqrt{2} \approx 1.41\).
6.7 5.7 The Geometric distribution
The geometric models the number of failures before the first success in a sequence of independent Bernoulli trials.
NoteDefinition: Geometric distribution
Let independent Bernoulli trials be performed with success probability \(p \in (0, 1)\). Let \(X\) be the number of failures before the first success. We write \(X \sim \mathrm{Geom}(p)\).
\[ p(k) = P(X = k) = (1 - p)^k\, p, \qquad k = 0, 1, 2, \ldots \]
With \(q = 1 - p\), the mean and variance are
\[ \mu = \frac{q}{p} = \frac{1 - p}{p}, \qquad \sigma^2 = \frac{q}{p^2} = \frac{1 - p}{p^2}. \]
WarningTwo conventions
Some books count the trial number of the first success instead, \(Y = X + 1\), with \(P(Y = k) = (1 - p)^{k - 1} p\) for \(k = 1, 2, \ldots\) and \(\mathbb{E}[Y] = 1/p\). R’s dgeom uses the “failures before first success” convention (i.e., \(X\), starting at 0). Always check which one a source assumes.
6.7.1 5.7.1 The memoryless property
The geometric distribution is the only discrete distribution with the memoryless property:
\[ P(X \geq s + t \mid X \geq t) = P(X \geq s), \qquad s, t \geq 0. \]
Past failures contain no information about the future: every trial is identical regardless of history. This is sometimes a useful normative reminder (“don’t get discouraged — each call is a fresh trial”) but also a strong modelling limitation. In reality, fatigue, learning effects, or selection on order (good leads called first) often make later trials behave differently.
6.7.2 5.7.2 Worked example: door-to-door sales
A salesman has a 20% chance of making a sale at each house (\(p = 0.2\)); let \(X\) be the number of failures before the first sale. Then \(X \sim \mathrm{Geom}(0.2)\).
- \(\mathbb{E}[X] = \frac{0.8}{0.2} = 4\) failures on average. Equivalently, the first sale is expected on the 5th visit (\(\mathbb{E}[Y] = 5\)).
- \(P(X = 0) = (0.8)^0 \times 0.2 = 0.2\): 20% chance the very first house is a sale.
- \(P(X > 4) = (0.8)^5 = 0.3277\), so \(P(X \leq 4) = 0.6723\) — about a 67% chance of a sale within five visits.
6.7.3 5.7.3 Business example: startup fundraising
A founder pitches to venture-capital firms; each pitch has \(p = 0.08\) of receiving an offer. Let \(X\) be the number of rejections before the first offer, so \(X \sim \mathrm{Geom}(0.08)\).
- Expected rejections: \(\mathbb{E}[X] = 0.92/0.08 = 11.5\).
- Funded within 3 pitches: \(P(Y \leq 3) = 1 - (0.92)^3 = 0.2213\).
- More than 10 pitches needed: \(P(Y > 10) = (0.92)^{10} = 0.4344\).
By the memoryless property, the probability of being funded in the next 3 pitches given 8 prior rejections is still \(0.2213\) — the 8 rejections carry no information.
6.8 5.8 The Negative Binomial (brief mention)
A natural generalisation of the geometric asks: how many failures occur before the \(r\)-th success?
NoteDefinition: Negative Binomial distribution
Let independent Bernoulli trials be performed with success probability \(p\), and let \(X\) be the number of failures before the \(r\)-th success. We write \(X \sim \mathrm{NB}(r, p)\).
\[ p(k) = \binom{k + r - 1}{k} p^r (1 - p)^k, \qquad k = 0, 1, 2, \ldots \]
with mean \(\mu = r(1 - p)/p\) and variance \(\sigma^2 = r(1 - p)/p^2\). When \(r = 1\) this reduces to the geometric. Continuing the startup example, if three independent offers are needed for a full funding round, then \(X \sim \mathrm{NB}(3, 0.08)\) with \(\mathbb{E}[X] = 34.5\) rejections expected.
6.9 5.9 Identifying the right distribution
The most important applied skill in this chapter is recognising which distribution to use. The questions below — answered in order — work for every scenario in this course.
- Are all outcomes equally likely? → Discrete Uniform.
- Is it a single yes/no trial? → Bernoulli\((p)\).
- Are we counting successes in \(n\) fixed trials?
- With replacement / large population: Binomial\((n, p)\).
- Without replacement from finite population: Hypergeometric\((N, K, n)\).
- Are we counting events in a continuous interval at a constant rate? → Poisson\((\lambda)\).
- Are we counting trials until the first success? → Geometric\((p)\).
- Trials until the \(r\)-th success? → Negative Binomial\((r, p)\).
| Distribution | PMF | Mean \(\mu\) | Variance \(\sigma^2\) |
|---|---|---|---|
| Bernoulli\((p)\) | \(p^x (1 - p)^{1 - x}\) | \(p\) | \(p(1 - p)\) |
| Binomial\((n, p)\) | \(\binom{n}{k} p^k (1 - p)^{n - k}\) | \(np\) | \(np(1 - p)\) |
| Hyp\((N, K, n)\) | \(\dfrac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}\) | \(n\dfrac{K}{N}\) | \(n\dfrac{K}{N}\dfrac{N - K}{N}\dfrac{N - n}{N - 1}\) |
| Poisson\((\lambda)\) | \(\dfrac{e^{-\lambda}\lambda^k}{k!}\) | \(\lambda\) | \(\lambda\) |
| Geom\((p)\) | \((1 - p)^k\, p\) | \(\dfrac{1 - p}{p}\) | \(\dfrac{1 - p}{p^2}\) |
6.10 5.10 R Lab — PMFs, CDFs and the chocolate factory
The running theme of this lab is quality control at a chocolate factory. Each section introduces a distribution through a realistic scenario.
6.10.1 5.10.1 Binomial: the d/p/q/r family
A chocolate factory has a 5% defect rate. An inspector samples 20 bars: \(X \sim B(20, 0.05)\).
Code
n <- 20
p <- 0.05
# Point probabilities P(X = 0), P(X = 1), P(X = 2)
dbinom(0:2, size = n, prob = p)[1] 0.3584859 0.3773536 0.1886768Code
# Cumulative P(X <= 2)
pbinom(2, size = n, prob = p)[1] 0.9245163Code
# Upper tail P(X > 3) via complement
1 - pbinom(3, size = n, prob = p)[1] 0.01590153There is roughly a 92.5% chance that at most two bars are defective and under 2% chance of more than three defectives.
Code
x_vals <- 0:n
probs <- dbinom(x_vals, size = n, prob = p)
barplot(probs, names.arg = x_vals, col = "chocolate3", border = "white",
main = "Binomial PMF: B(20, 0.05)",
xlab = "Number of defective bars", ylab = "Probability")
The distribution is heavily right-skewed because the defect rate is low.
Code
mu_b <- n * p
sigma_b <- sqrt(n * p * (1 - p))
cat("E[X] =", mu_b, " SD[X] =", round(sigma_b, 3), "\n")E[X] = 1 SD[X] = 0.975 6.10.2 5.10.2 Binomial shape across \(p\)
The shape of \(B(n, p)\) changes dramatically with \(p\). Side-by-side PMFs for \(n = 10\):
Code
op <- par(mfrow = c(1, 3))
for (pp in c(0.2, 0.5, 0.8)) {
barplot(dbinom(0:10, size = 10, prob = pp),
names.arg = 0:10, col = "steelblue", border = "white",
main = paste0("B(10, ", pp, ")"),
xlab = "k", ylab = "P(X = k)")
}
Code
par(op)Low \(p\) gives right-skew, \(p = 0.5\) is symmetric, high \(p\) gives left-skew.
6.10.3 5.10.3 Poisson
The customer-service desk receives an average of 3 complaints per day: \(Y \sim \mathrm{Poisson}(3)\).
Code
lambda <- 3
dpois(0:5, lambda) # P(Y = 0), ..., P(Y = 5)[1] 0.04978707 0.14936121 0.22404181 0.22404181 0.16803136 0.10081881Code
ppois(5, lambda) # P(Y <= 5)[1] 0.9160821How the shape changes with \(\lambda\):
Code
y_vals <- 0:12
op <- par(mfrow = c(1, 3))
for (lam in c(1, 3, 8)) {
barplot(dpois(y_vals, lam), names.arg = y_vals,
col = "steelblue", border = "white",
main = paste0("Poisson(", lam, ")"),
xlab = "y", ylab = "P(Y = y)")
}
Code
par(op)For small \(\lambda\) the distribution is right-skewed; as \(\lambda\) grows it becomes nearly symmetric — an illustration of the CLT applied to count data.
6.10.4 5.10.4 Hypergeometric
Two scenarios.
Code
# Loteria 6/49: 6 winning numbers among 49; you pick 6
# dhyper(x, m, n, k): m = successes in urn, n = failures, k = draws
dhyper(0:6, m = 6, n = 43, k = 6)[1] 4.359650e-01 4.130195e-01 1.323780e-01 1.765040e-02 9.686197e-04
[6] 1.844990e-05 7.151124e-08The jackpot probability is about \(7.15 \times 10^{-8}\) (the last entry above).
Code
# Crate of 50 boxes, 5 defective; sample 10 without replacement
probs_h <- dhyper(0:5, m = 5, n = 45, k = 10)
names(probs_h) <- 0:5
round(probs_h, 4) 0 1 2 3 4 5
0.3106 0.4313 0.2098 0.0442 0.0040 0.0001 Code
barplot(probs_h, col = "tomato", border = "white",
main = "Hypergeometric: 5 defective in 50, sample 10",
xlab = "Defectives found", ylab = "Probability")
6.10.5 5.10.5 Geometric and the memoryless property
A salesperson closes a deal with probability \(p = 0.20\) per visit. Let \(W\) be the number of failures before the first success: \(W \sim \mathrm{Geom}(0.20)\) in R’s convention.
Code
p_sale <- 0.20
# First sale on the 4th visit => 3 failures then 1 success
dgeom(3, prob = p_sale)[1] 0.1024Code
# P(need more than 5 visits) => P(W > 4)
1 - pgeom(4, prob = p_sale)[1] 0.32768Verify the memoryless property numerically:
Code
s <- 3; t <- 2
cond <- (1 - pgeom(s + t, p_sale)) / (1 - pgeom(s, p_sale))
uncond <- 1 - pgeom(t, p_sale)
cat("P(W >", s + t, "| W >", s, ") =", round(cond, 4), "\n")P(W > 5 | W > 3 ) = 0.64 Code
cat("P(W >", t, ") =", round(uncond, 4), "\n")P(W > 2 ) = 0.512 The two numbers are identical: past failures contain no information about the future.
6.10.6 5.10.6 Binomial–Poisson approximation, visualised
Set \(n = 1000\), \(p = 0.002\), so \(\lambda = np = 2\).
Code
n2 <- 1000
p2 <- 0.002
lam2 <- n2 * p2
x_range <- 0:10
binom_probs <- dbinom(x_range, size = n2, prob = p2)
pois_probs <- dpois(x_range, lambda = lam2)
barplot(rbind(binom_probs, pois_probs), beside = TRUE,
names.arg = x_range, col = c("chocolate3", "steelblue"),
border = "white",
main = "B(1000, 0.002) vs Pois(2)",
xlab = "x", ylab = "Probability")
legend("topright", legend = c("Binomial", "Poisson"),
fill = c("chocolate3", "steelblue"), bty = "n")
Code
round(data.frame(x = x_range, Binomial = binom_probs,
Poisson = pois_probs,
Diff = binom_probs - pois_probs), 5) x Binomial Poisson Diff
1 0 0.13506 0.13534 -0.00027
2 1 0.27067 0.27067 0.00000
3 2 0.27094 0.27067 0.00027
4 3 0.18063 0.18045 0.00018
5 4 0.09022 0.09022 0.00000
6 5 0.03602 0.03609 -0.00007
7 6 0.01197 0.01203 -0.00006
8 7 0.00341 0.00344 -0.00003
9 8 0.00085 0.00086 -0.00001
10 9 0.00019 0.00019 0.00000
11 10 0.00004 0.00004 0.00000The maximum difference is on the order of \(10^{-4}\) — visually the two PMFs are indistinguishable.
Self-check
TipQ1. Binomial PMF
\(X \sim B(n, p)\) counts the number of successes in \(n\) independent Bernoulli trials with constant probability \(p\). The PMF is:
- A. \(P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}\) for \(k = 0, 1, \ldots, n\).
- B. \(P(X = k) = e^{-p} p^k / k!\).
- C. \(P(X = k) = (1 - p)^{k - 1} p\).
- D. \(P(X = k) = p^k\) for any \(k\).
Answer: A. Option B is the Poisson, C is the geometric, D is not a valid PMF.
TipQ2. Binomial mean and variance
If \(X \sim B(n, p)\), then \(\mathbb{E}[X]\) and \(\operatorname{Var}(X)\) equal:
- A. \(\mathbb{E}[X] = np\) and \(\operatorname{Var}(X) = np(1 - p)\).
- B. \(\mathbb{E}[X] = p\) and \(\operatorname{Var}(X) = p(1 - p)\).
- C. \(\mathbb{E}[X] = np^2\) and \(\operatorname{Var}(X) = np\).
- D. \(\mathbb{E}[X] = n\) and \(\operatorname{Var}(X) = n^2 p\).
Answer: A. B describes a single Bernoulli trial, not the sum of \(n\) of them.
TipQ3. Reading R output
In R, pbinom(2, size = 20, prob = 0.05) returns:
- A. \(P(X \leq 2)\), the cumulative probability up to and including 2.
- B. \(P(X = 2)\), the point probability at 2.
- C. \(P(X > 2)\), the upper-tail probability.
- D. The second quantile of the distribution.
Answer: A. dbinom returns the point probability, pbinom the cumulative, qbinom the quantile.
TipQ4. Poisson PMF
The Poisson PMF with rate \(\lambda > 0\) is:
- A. \(P(Y = k) = \dfrac{e^{-\lambda} \lambda^k}{k!}\) for \(k = 0, 1, 2, \ldots\).
- B. \(P(Y = k) = \binom{n}{k} \lambda^k\).
- C. \(P(Y = k) = (1 - \lambda)^{k - 1} \lambda\).
- D. \(P(Y = k) = \lambda^k / k\).
Answer: A. The Poisson has support on all non-negative integers; \(k!\) in the denominator is essential.
TipQ5. Defining feature of the Poisson
A defining feature of the Poisson distribution is:
- A. \(\mathbb{E}[Y] = \operatorname{Var}(Y) = \lambda\) (equidispersion).
- B. \(\mathbb{E}[Y] = \lambda\) but \(\operatorname{Var}(Y) = \lambda^2\).
- C. \(\mathbb{E}[Y] = 0\) for every \(\lambda\).
- D. \(\operatorname{Var}(Y) = \mathbb{E}[Y] / 2\).
Answer: A. Empirical data with variance much larger than the mean (“overdispersion”) signal that Poisson is the wrong model.
TipQ6. Poisson approximation
When \(n\) is large and \(p\) is small, \(B(n, p)\) is well approximated by:
- A. \(\mathrm{Poisson}(\lambda = np)\).
- B. \(\mathrm{Poisson}(\lambda = n - p)\).
- C. \(N(np,\, np(1 - p))\).
- D. \(\mathrm{Geom}(p)\).
Answer: A. The rate parameter is \(\lambda = np\); rules of thumb are \(n \geq 30\), \(p \leq 0.10\).
TipQ7. When to use the hypergeometric
The hypergeometric distribution arises when:
- A. We draw \(k\) items without replacement from a finite population of size \(N\) containing \(m\) successes.
- B. We perform infinitely many independent trials.
- C. We count events in continuous time.
- D. We count failures before the first success.
Answer: A. Without replacement is the key signal. With replacement (or large \(N\)) leads to the binomial; counting in time leads to Poisson.
TipQ8. Memoryless property
The geometric distribution has the memoryless property: \(P(W > s + t \mid W > s) = P(W > t)\). Intuitively this means:
- A. Past failures contain no information about how many more failures remain — each trial is a fresh trial with the same success probability \(p\).
- B. Past failures make future success more likely (gambler’s fallacy).
- C. Past failures make future success less likely.
- D. \(W\) is bounded above by some fixed number.
Answer: A. This is sometimes a useful normative reminder, sometimes a strong modelling limitation when fatigue or learning effects matter.
Exercises
6.10.7 Exercise 5.1 ★ — Discrete uniform: fair die
A fair six-sided die is rolled once; let \(X\) be the number on the upper face.
- Write the PMF of \(X\).
- Compute \(P(X > 4)\).
- Compute \(\mathbb{E}[X]\) and \(\operatorname{Var}(X)\).
TipSolution
\(P(X = k) = 1/6\) for \(k \in \{1, 2, 3, 4, 5, 6\}\).
\(P(X > 4) = P(X = 5) + P(X = 6) = 2/6 = 1/3 \approx 0.3333\).
\(\mathbb{E}[X] = (6 + 1)/2 = 3.5\); \(\operatorname{Var}(X) = (36 - 1)/12 = 35/12 \approx 2.917\).
6.10.8 Exercise 5.2 ★ — Binomial: defect rate
A factory produces electronic components with a defect rate of 4%. A random sample of 15 components is selected; let \(X\) be the number of defective components.
- State the distribution of \(X\) and its parameters.
- Compute \(P(X = 2)\).
- Compute \(P(X \leq 1)\).
- Find \(\mathbb{E}[X]\) and \(\operatorname{Var}(X)\).
TipSolution
\(X \sim B(n = 15, p = 0.04)\).
\(P(X = 2) = \binom{15}{2}(0.04)^2 (0.96)^{13} = 105 \times 0.0016 \times 0.5882 \approx 0.0988\).
\(P(X = 0) = (0.96)^{15} \approx 0.5421\); \(P(X = 1) = 15 \times 0.04 \times (0.96)^{14} \approx 0.3388\); so \(P(X \leq 1) \approx 0.8809\).
\(\mathbb{E}[X] = 0.60\); \(\operatorname{Var}(X) = 15 \times 0.04 \times 0.96 = 0.576\).
6.10.9 Exercise 5.3 ★ — Poisson: emergency room
A hospital emergency room receives an average of 5 patients per hour.
- State the distribution of \(X\).
- Compute \(P(X = 3)\).
- Compute \(P(X \geq 2)\).
- Compute \(P(X = 0)\) and interpret.
TipSolution
\(X \sim \mathrm{Poisson}(\lambda = 5)\).
\(P(X = 3) = e^{-5} \cdot 5^3 / 3! \approx 0.1404\).
\(P(X \geq 2) = 1 - [e^{-5} + 5 e^{-5}] = 1 - 0.04043 \approx 0.9596\).
\(P(X = 0) = e^{-5} \approx 0.0067\) — only a 0.67% chance of a patient-free hour, consistent with a busy ER.
6.10.10 Exercise 5.4 ★★ — Binomial cumulative: online returns
An online store has an 8% return rate. In a batch of 25 shipped orders, let \(X\) be the number of returns.
- State the distribution of \(X\).
- Compute \(P(X > 3)\).
- Compute \(P(X \leq 1)\).
6.10.11 Exercise 5.5 ★★ — Poisson approximation: spam emails
A company receives 500 emails per day, each spam with probability 0.2% independently. Let \(X\) be the daily number of spam emails.
- State the exact distribution of \(X\).
- Justify a Poisson approximation and give \(\lambda\).
- Using the approximation, compute \(P(X \geq 3)\).
6.10.12 Exercise 5.6 ★★ — Hypergeometric: audit selection
A department has 15 expense reports, of which 5 contain irregularities. An auditor randomly selects 4 reports (without replacement).
- Compute \(P(X = 2)\).
- Compute \(P(X = 0)\).
- Find \(\mathbb{E}[X]\).
- How different would part (a) be if we used a binomial approximation? Why is the answer different here than in a “large population” setting?
6.10.13 Exercise 5.7 ★★ — Geometric: sales visits
A salesperson has a 15% probability of making a sale on each independent visit. Let \(X\) be the number of visits needed to make the first sale.
- State the distribution and explain which convention (\(X\) = trial of first success, or \(X\) = failures before first success) you are using.
- Compute \(P(X = 4)\).
- Compute \(\mathbb{E}[X]\).
- Compute \(P(X > 5)\).
6.10.14 Exercise 5.8 ★★★ — Quality control: full problem
A pharmaceutical company produces batches of 200 tablets; 2% of tablets are under-dosed. An inspector samples 20 tablets from each batch.
- Justify why the binomial is a reasonable model for the number of under-dosed tablets \(X\), even though the underlying sampling is without replacement.
- Compute \(P(X = 0)\) and \(P(X \geq 2)\).
- The batch is rejected if \(X \geq 2\). What is the rejection probability?
- Management wants the rejection probability to be at most 5%. Find (numerically) the maximum defect rate \(p^\*\) that achieves this, keeping \(n = 20\) and the same rejection rule.