Question bank — Binomial distribution — PMF, mean, variance

False. The count $X$ is always an integer, but its average $np$ need not be. Example: $n=5,p=0.5$ gives mean $2.5$ — an average nobody ever actually observes. ::: False — $np$ is an average, so values like $2.5$ are fine even though each individual outcome is a whole number of successes.

$X\sim \text{Bin}(n,p)$ then $n-X\sim \text{Bin}(n,1-p)$. ::: True — "number of successes" and "number of failures" swap roles; failures happen with probability $1-p$ on the same $n$ independent trials, so $n-X$ is binomial with success prob $1-p$.

$n$ doubles both the mean and the standard deviation. ::: False — the mean $np$ doubles, but SD is $\sqrt{np(1-p)}$, so doubling $n$ multiplies SD by only $\sqrt{2}$. Spread grows slower than the mean, which is why large samples get proportionally "tighter."

$n$, variance is largest exactly when the mean is largest. ::: False — the mean $np$ grows all the way to $n$ as $p\to 1$, but variance $np(1-p)$ peaks at $p=0.5$ and collapses to $0$ at $p=1$. Maximum spread and maximum mean live at different $p$.

$p$ is small. ::: False — $X$ counts successes, so it lives on $\{0,1,\dots ,n\}$ regardless of $p$. A small $p$ just piles probability near $0$, never below it.

$\text{Bin}(n_1,p)+\text{Bin}(n_2,p)$ gives $\text{Bin}(n_1+n_2,p)$. ::: True — but only because they share the same $p$. Then it's just one big pool of $n_1+n_2$ independent trials with common success prob $p$. With different $p$'s the sum is not binomial at all.

$P(X=k)$ and $P(X=n-k)$ are always equal for a binomial. ::: False in general — they are equal only when $p=0.5$, where the distribution is symmetric. For $p\ne 0.5$ the factor $p^k(1-p{)}^{n-k}$ is unbalanced, so the two probabilities differ.

$X$ is always the one nearest $np$. ::: True (with a caveat) — the mode is $\lfloor (n+1)p\rfloor$, which sits at or right next to $np$. When $(n+1)p$ is a whole number there are two adjacent equally-likely modes.

$p=0.7$, the chance of passing all four is $4\times 0.7=2.8$." ::: Probabilities never exceed $1$, so $2.8$ is nonsense. You multiply independent probabilities, not add: $0.7^4\approx 0.24$. Adding is the error of confusing "all of" (AND → product) with expected count.

$P(X=3)=p^3(1-p{)}^{n-3}$ because that's the probability of three successes then failures." ::: That's the probability of one specific ordering. Many orderings give exactly 3 successes, so you must multiply by $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ to count them all. Dropping $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the single most common PMF mistake.

$(X)=np(1-p)$, so the standard deviation is $np(1-p)$ too." ::: SD is the square root of variance: $\sigma =\sqrt{np(1-p)}$. Reporting the variance itself as SD confuses the two — and gives the wrong units of spread.

$\text{Bin}(52,\frac{4}{52})$ for the number of aces in a 5-card poker hand." ::: Drawing without replacement means $p$ changes after each card, so trials are dependent — this is hypergeometric, not binomial. Also the number of trials is $5$ draws, not $52$.

$np(1-p)$ because both mean and variance are built from $np$." ::: The mean is $np$ with no $(1-p)$ factor — it comes from linearity of expectation, where each trial adds $p$ on average. Only the variance carries the extra $q=1-p$.

$P(X=1)+P(X=2)+\cdots +P(X=n)$; there's no shortcut." ::: There is: $P(X\ge 1)=1-P(X=0)=1-(1-p{)}^n$. The complement replaces a long sum with one subtraction. Missing this is a work-efficiency trap, not a correctness one.

$E[X]=np$ only holds when trials are independent." ::: Linearity of expectation holds even for dependent variables, so $E[X]=np$ needs no independence. It's the variance formula $np(1-p)$ that requires independence (so covariances vanish).

::: $E[\sum X_i]=\sum E[X_i]$ always (linearity). But $\text{Var}(\sum X_i)=\sum \text{Var}(X_i)+{\sum }_{i\ne j}\text{Cov}(X_i,X_j)$; only independence forces every covariance to $0$ so the variances simply add.

$p=0.5$? ::: $p(1-p)$ is a downward parabola in $p$, peaking at $p=0.5$. Intuitively, $p=0.5$ is maximal uncertainty — you truly cannot predict a single trial — while $p=0$ or $1$ are certain, so their spread is zero.

$\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ appear, and why exactly that formula? ::: It counts how many arrangements of $k$ successes among $n$ ordered trials exist; each arrangement has the identical probability $p^k(1-p{)}^{n-k}$, so we multiply by the count. $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ is the number of ways to choose which slots succeed.

$1$? ::: Summing $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ over all $k$ is the Binomial Theorem expansion of $(p+(1-p){)}^n=1^n=1$. The algebra of the binomial expansion is literally the same bookkeeping as the probabilities.

$p(1-p)$ and not something with a square root? ::: For an indicator $X_i$, $X_i^2=X_i$, so $E[X_i^2]=p$ and $\text{Var}=E[X_i^2]-(E[X_i]{)}^2=p-p^2=p(1-p)$. The shortcut $X_i^2=X_i$ is what makes it so clean.

$n$? ::: $X$ is a sum of many independent identical Bernoullis, so the Central Limit Theorem pulls its shape toward a bell curve — the Normal approximation to binomial with matching mean $np$ and variance $np(1-p)$.

::: When $n\to \infty$ and $p\to 0$ with $np=\lambda$ held fixed, the $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ formula limits to $\frac{{\lambda }^k{e}^{-\lambda }}{k!}$ — the Poisson distribution. It describes counts of rare successes over many chances.

$\text{Bin}(1,p)$ reduce to? ::: A single Bernoulli trial: $X$ is $1$ with prob $p$ and $0$ with prob $1-p$. Mean $p$, variance $p(1-p)$ — the atomic building block of every binomial.

$p=0$? ::: $X$ is $0$ with certainty — no trial can succeed. Mean $=n\cdot 0=0$ and variance $=n\cdot 0\cdot 1=0$; the "randomness" has vanished entirely.

$p=1$? ::: $X=n$ with certainty — every trial succeeds. Mean $=n$, variance $=n\cdot 1\cdot 0=0$. Again a degenerate, spike distribution with no spread.

$P(X=0)$ and $P(X=n)$ in general? ::: $P(X=0)=(1-p{)}^n$ (all failures) and $P(X=n)=p^n$ (all successes), since $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=\left(\genfrac{}{}{0ex}{}{n}{n}\right)=1$ leaves a single ordering each.

$\text{Bin}(0,p)$ — zero trials? ::: The only possible value is $X=0$, with probability $1$. Mean and variance are both $0$; there is nothing to count. A valid but trivial degenerate case.

$(n,p)$ pairs give the same mean but wildly different spread? ::: Yes — $\text{Bin}(100,0.05)$ and $\text{Bin}(5,1.0)$ both have mean $5$, but the first has variance $4.75$ while the second has variance $0$. Equal means say nothing about shape.