Probability, randomness, and the discrete distributions that show up everywhere in ecology

Sample space: the four equally likely birth orders are {BB, BG, GB, GG}, each with probability 1/4.

(a) "At least one girl" rules out BB. We are left with {BG, GB, GG}, three equally likely outcomes. Only GG is "both girls". So $P(\text{both} \mid \text{at least one girl}) = \tfrac{1}{3}$.

(b) "Older child is a girl" rules out BB and GB. We are left with {BG, GG}, two equally likely outcomes. So $P(\text{both} \mid \text{older is girl}) = \tfrac{1}{2}$.

(c) Knowing which child is a girl (a more specific piece of information) shrinks the sample space more than just knowing one of them is.

Let $B$ = "taxi is blue", $W_B$ = "witness says blue".

• $P(B) = 0.01$, $P(\bar B) = 0.99$
• $P(W_B \mid B) = 0.99$
• $P(W_B \mid \bar B) = 0.05$ (the 5% misclassification of green)

Total probability of "says blue":

Bayes:

So only ~17%, even though the witness is "99% reliable" on blue taxis.

Where the small answer really comes from: the odds form. Don't stop at the plug-in; the same problem read in odds shows exactly why a "99% reliable" witness still loses. Divide Bayes' rule for $B$ by Bayes' rule for $\bar B$: the shared denominator $P(W_B)$ cancels, and you are left with a rule worth memorizing:

Posterior odds = prior odds × likelihood ratio. The evidence only ever multiplies the odds you started with. Here the prior odds are $1:99$ and the likelihood ratio is $0.99/0.05 = 19.8$, so the posterior odds are $19.8/99 = 0.2$, i.e. $1:5$, which is $P = \tfrac{1}{1+5} = \tfrac{1}{6} \approx 0.167$, the same answer, now with the mechanism exposed. To reach even money (odds $1:1$) the test would need a likelihood ratio near 99, it would have to be about five times sharper than it is, just to overcome a base rate that starts you 99-to-1 in the hole. That is the general lesson. Against a rare event, the base rate sets the odds you must climb out of, and even a very accurate test barely dents it.

p_B <- 0.01; p_W_B <- 0.99; p_W_notB <- 0.05
(p_W_B * p_B) / (p_W_B * p_B + p_W_notB * (1 - p_B)) # 0.1667

Let $X_i$ be 1 if person $i$ is shorter than both neighbors, else 0. The total count is $X = \sum X_i$. By linearity of expectation, $E[X] = \sum E[X_i] = n \cdot P(\text{person 1 shorter than both neighbors})$.

Consider person $i$ together with their two neighbors. Because everyone is seated in random order and all heights are distinct, each of the $3! = 6$ height orderings of these three people is equally likely. Person $i$ is shorter than both neighbors exactly when $i$ is the smallest of the three. Fixing $i$ as the smallest leaves the two neighbors free to be arranged in $2! = 2$ ways, so 2 of the 6 orderings qualify. Hence $P(\text{person } i \text{ shorter than both neighbors}) = 2/6 = 1/3$.

Therefore $E[X] = n/3$.

Punnett: CC : Cc : Cc : cc = 1:1:1:1, each with probability 1/4. Unaffected = CC, Cc, or Cc.

(a) $P(\text{unaffected}) = 1/4 + 1/4 + 1/4 = 3/4$. (Or just $1 - P(cc) = 1 - 1/4 = 3/4$.)

(b) Children are independent: $(3/4)^4 = 81/256 \approx 0.316$.

(3/4)^4 # 0.3164
dbinom(0, size=4, prob=1/4) # same thing

Theoretical values. Every entry in this table is a formula evaluated exactly, none of it is simulation output. You can fill the whole table in before you draw a single number, and doing so first is the point:

Now compare your run, and mind the gap. Your 1000 draws will not reproduce this table. With set.seed(2025) the NB($k=0.5$) row comes back with a sample variance of $244.63$ against the theoretical $210$, and $P(Y=0)=0.202$ against $0.218$; the Poisson row’s sample variance, by contrast, lands almost on top of its theoretical variance ($9.49$ against $10$). Nothing has gone wrong. A sample variance is itself a random quantity, and for a heavy right tail it is a very noisy one. A handful of extreme transects shift it by tens of units, so 1000 draws pin it down only loosely. The rule to carry forward is that the more clumped the model, the more data you need before an empirical summary can be trusted. When a real dataset's variance sits this far from what your model predicts, the first question is “do I have enough data to tell?”, not “is the model wrong?”

The variance column. Every row has mean 10 by construction; the story is entirely in the variance. Start from what the NB literally is, a Poisson whose rate is itself random (gamma-distributed) across transects, the mathematical version of "animals clump." That mixing gives the mean-variance law

Read it as two pieces: the $\mu$ is the irreducible Poisson noise, and the $\mu^2/k$ is the extra variance clumping adds. That second term is the whole point. It is what a plain Poisson cannot produce. The general move is to see $k$ as a clumping dial: as $k \to \infty$ the extra term vanishes and you recover $\mathrm{Var}=\mu$ (the Poisson is just NB with the clumping turned off), and as $k$ shrinks the extra term explodes. Only now plug in, with $\mu=10$:

Halving $k$ from 1 to 0.5 doubles the excess term (100 → 200), so clumping bites hardest at small $k$.

The $P(Y=0)$ column. A zero transect means "not one whale," and the NB pmf evaluated at zero collapses to a clean closed form,

so the same clumping dial controls how common empty transects are. Plug in $\mu=10$: $k=5$ gives $(5/15)^5=(1/3)^5=1/243\approx 0.004$; $k=1$ gives $(1/11)^1\approx 0.091$; $k=0.5$ gives $(0.5/10.5)^{0.5}=\sqrt{1/21}\approx 0.218$. The Poisson is the limit again: $(k/(k+\mu))^k \to e^{-\mu}$ as $k\to\infty$, and $e^{-10}\approx 0.0000454$, which rounds to the $0.000$ in the table. So a Poisson essentially never gives an empty transect, while NB($k=0.5$) gives one more than a fifth of the time.

(d) Small $k$ inflates both tails at once: the $P(Y=0)$ and $P(Y>20)$ columns rise together, which is exactly the "many zeros, a few huge counts" signature. The dataset described matches NB with small $k$; most ecological count data are like this.

Conceptual takeaway: the Poisson is a one-parameter distribution and is forced to have variance equal to its mean. Real ecological counts almost never satisfy that. Use the negative binomial unless you have a specific reason not to.

k <- 0:5
probs <- dbinom(k, size=5, prob=0.7)
round(probs, 3)
# 0.002 0.028 0.132 0.309 0.360 0.168
barplot(probs, names.arg=k, ylab="P(survivors)")

This is the Binomial distribution, $X \sim \text{Bin}(n=5, p=0.7)$. Each marten either survives or does not, independently and with the same probability, so each one is a Bernoulli($p$) trial. The number of survivors is the sum of five independent Bernoulli trials, and a sum of independent Bernoulli trials is exactly what Binomial means:

(a) $Y \sim \text{Bin}(n=24, p=0.12)$.

dbinom(3, 24, 0.12) # P(Y=3) = 0.2387
pbinom(4, 24, 0.12) # P(Y<=4) = 0.8471
dbinom(0, 24, 0.12) # P(Y=0) = 0.0465

(b) Same pmf as Exercise A2, just bigger numbers: $P(Y=3)=\binom{24}{3}(0.12)^3(0.88)^{21}=2024\times0.001728\times0.0683=0.239$.

(c) Cumulative rather than point: $P(Y \le 4) = \sum_{k=0}^{4} P(Y=k)$, which R adds up for you: pbinom(4, 24, 0.12) $= 0.847$. So at 12% prevalence, "four or fewer infected in a sample of 24" is the ordinary result. It happens about 85% of the time.

(d) The zero-case probability is the one you can do in your head, because at $k=0$ the combinatorial factor is $\binom{24}{0}=1$ and the whole pmf collapses to "every single deer must dodge infection":

(e) That is the load-bearing quantity for surveillance, so make it general. The chance of catching at least one case in $n$ samples is its complement,

and flipping it around tells a program how many deer to sample. To be $95\%$ sure of detecting disease at prevalence $p$, solve $1-(1-p)^n \ge 0.95$ for $n$. Here, with $p=0.12$, the inequality rearranges to $n \ge \log(0.05)/\log(0.88) = 23.44$, so $n=24$ is the smallest sample that clears the bar, and it clears it barely: $P(Y\ge1)=1-0.0465=0.9535$, against $0.9471$ at $n=23$. Even at that prevalence, a smaller sample would leave a real chance of a clean-but-wrong "no cases found." This is precisely the arithmetic behind "freedom from disease" sample-size calculations.

p <- c(0.20, 0.45, 0.25, 0.10)
dmultinom(c(18, 38, 18, 6), prob=p) # ~0.0011
dmultinom(c(16, 36, 20, 8), prob=p) # the modal flight: 0.001832

set.seed(1)
flights <- rmultinom(5000, size=80, prob=p)
hist(flights[3,], main="Calves out of 80", xlab="count")

(a) Where the 0.0011 comes from. The multinomial pmf is the binomial's "many boxes" generalization: one multinomial coefficient counting the orderings, times each category's probability raised to its observed count,

With $\mathbf{x}=(18,38,18,6)$, $n=80$, that is $\dfrac{80!}{18!\,38!\,18!\,6!}\,(0.20)^{18}(0.45)^{38}(0.25)^{18}(0.10)^{6}\approx 0.0011$. It reads small, but every specific outcome is small, because there are thousands of plausible flights, and the probability has to be shared out among all of them. Search the whole support and the modal flight turns out to be $(16, 36, 20, 8)$, exactly $n\mathbf{p}$, each category at its own expected count, and even that single most likely outcome carries only $0.001832$, under $0.2\%$ of the probability. Our $(18,38,18,6)$, at $0.00117$, is a perfectly ordinary flight.

(c) Why each margin is Binomial: the collapse trick. Don't take it on faith; derive it, because it is the move that lets you reason about one category at a time. Fix your attention on calves and lump the other three classes into a single "not-calf" box. Each of the 80 elk is now just calf (prob $0.25$) or not-calf (prob $0.20+0.45+0.10=0.75$), independently, and "count of successes in $n$ independent yes/no trials with fixed $p$" is the definition of the Binomial. So

The general rule: any single category of a Multinomial$(n,\mathbf{p})$ is Binomial$(n,p_i)$. Collapse the rest into "other" and the multi-box problem becomes the two-box problem you already know. (Note $\mathrm{Var}=15$ is below the mean 20. That is not special to the multinomial: any Binomial$(n,p)$ has $\mathrm{Var}/\mathrm{mean} = 1-p$, which is below 1 for every $p>0$, so a Binomial is always under-dispersed relative to a Poisson of the same mean. A multinomial margin inherits it because that margin is a Binomial. The reason is the fixed total. With $n$ pinned at 80, a category cannot run away the way an unbounded count can.)

p <- c(.30, .20, .15, .20, .15)
rmultinom(1, 80, p)
sims <- rmultinom(500, 80, p)
summary(sims[5,])

Same collapse trick as the elk (Exercise A4): lump the other four responses into "not strongly-support" and the marginal is $\text{Bin}(80, 0.15)$, with $E=np=12$ and $\mathrm{Var}=np(1-p)=80\cdot0.15\cdot0.85=10.2$, so $\mathrm{SD}=\sqrt{10.2}\approx 3.2$. That is the punchline made concrete. A typical single survey lands anywhere from about 6 to 18 "strongly support" votes (12 $\pm$ two SDs) even though the underlying rate never moved. The sample-to-sample range is large compared to the point estimate, which is exactly why one survey is weak evidence.