Question bank — Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial

True — $X=X_1+\cdots +X_n$ with each $X_i\in \{0,1\}$; this decomposition is exactly why $E[X]=np$ and $\mathrm{Var}(X)=npq$.

True — the only values are $0$ and $1$, and $0^2=0$, $1^2=1$; this is the trick that gives $E[X^2]=E[X]=p$ and hence $\mathrm{Var}=p-p^2$.

False — that equality is the signature of Poisson only (both $=\lambda$); Bernoulli has $p$ vs $pq$, Binomial $np$ vs $npq$, etc.

False — linearity holds for any variables, dependent or not; independence is only needed to add variances.

False in the trials-counting convention ($X\ge 1$, since you need at least one trial to get the first success); true in the failures-counting convention, where $0$ means "success on the very first trial, zero failures."

True — waiting for the "1st success" is the same as "trials to first success," and $\left(\genfrac{}{}{0ex}{}{k-1}{0}\right)=1$ collapses the formula to $q^{k-1}p$.

False — $k$ ranges over $0,1,2,\dots$ with no upper bound; it's the $n\to \infty$ limit of Binomial, so the ceiling $n$ disappears.

True — success is certain on trial 1, so $X=1$ always; $\mathrm{Var}=q/p^2=0/1=0$, a degenerate point mass.

False — in Geometric the success position is fixed (it's the last trial), so there is nothing to "choose"; no binomial coefficient appears.

True — $P(X>m+n\mid X>m)=P(X>n)$; already-waited trials give no information, which is memorylessness, unique to Geometric among discrete laws.

Wrong — the $k$-th trial is forced to be the stopping success, so only $r-1$ successes are free among the first $k-1$ trials: $\left(\genfrac{}{}{0ex}{}{k-1}{r-1}\right)$.

Wrong window — Poisson $\lambda$ must scale to the interval: for 2 minutes use $\lambda =6$, giving $\frac{6^5{e}^{-6}}{5!}$.

Missing the covariance — in general $\mathrm{Var}(X+Y)=\mathrm{Var}X+\mathrm{Var}Y+2\mathrm{Cov}(X,Y)$; the cross term vanishes only under independence (Figure 2 shows this in the relay picture).

Overstated — the mean $np$ follows from linearity alone and holds even for dependent trials; independence is only needed for $\mathrm{Var}=npq$.

Wrong — $npq=np(1-p)\to \lambda \cdot 1=\lambda$ because $np\to \lambda$ stays fixed and $(1-p)\to 1$; the variance survives as $\lambda$.

Wrong — it's a geometric series: ${\sum }_{k\ge 1}{q}^{k-1}p=p\cdot \frac{1}{1-q}=p/p=1$, a perfectly valid distribution.

Signs flipped — the formula is $\mathrm{Var}=E[X^2]-(E[X]{)}^2=p-p^2=pq$, which is non-negative as a variance must be.

That factor collects all the "no success" probability across $n$ tiny trials; it is the standard exponential limit, which is exactly where $e^{-\lambda }$ enters the Poisson formula (Figure 1 shows the curve converging).

Because waiting for the $r$-th success is a relay of waits: run to success 1, then restart the clock and run to success 2, and so on — $r$ independent Geometric legs laid end-to-end, so their average lengths simply stack: $r\cdot \frac{1}{p}=\frac{r}{p}$ (Figure 2).

We need $\sum k{q}^{k-1}$, and differentiating $\sum q^k=\frac{1}{1-q}$ term-by-term produces exactly $\sum k{q}^{k-1}=\frac{1}{(1-q{)}^2}$ — the derivative "pulls down" the $k$ we need.

The stopping rule is "reach the $r$-th success"; by definition you halt the instant the $r$-th success lands, so the final trial is that success — otherwise you'd have stopped earlier.

If a success happens a fraction $p$ of the time (e.g. $1/6$ for a die's six), you expect one success per $1/p$ trials — waiting $6$ rolls for a six matches everyday intuition.

The trials are independent, so all covariances are zero and variances add cleanly; drop independence and cross-covariance terms would appear.

"Trials to first success" counts the success itself (mean $1/p$); "failures before success" drops that one trial, shifting everything by one and giving mean $q/p$.

It reduces to a single Bernoulli$(p)$: $\left(\genfrac{}{}{0ex}{}{1}{k}\right)p^k{q}^{1-k}$ with $k\in \{0,1\}$ — the atom the whole family is built from.

$X=0$ with certainty (no success is ever possible); mean $np=0$ and variance $npq=0$, a degenerate spike at $0$.

$X=n$ with certainty (every trial succeeds); mean $np=n$ and variance $npq=n\cdot 1\cdot 0=0$, a degenerate spike at $n$ — the mirror image of the $p=0$ case.

The first success never arrives — $P(X<\infty )=0$, the total probability leaks out to infinity, so $X$ is not a proper finite-valued variable and the mean $1/p$ blows up to $\infty$. This is the boundary the formula $1/p$ is warning about.

With zero success chance the $r$-th success never comes for any finite $r\ge 1$; $X$ is infinite with probability $1$ (a defective distribution), and the mean $r/p$ diverges — the same leak as Geometric, amplified $r$-fold.

There is no fixed number of opportunities — in the limit $n\to \infty$, so any non-negative integer $k$ has positive probability, however large.

$X=0$ with certainty: $P(X=0)=0^0{e}^0/0!=1$ and $P(X=k)=0$ for $k\ge 1$; mean $=$ variance $=\lambda =0$, a degenerate spike at $0$ (no events ever occur).

$k=r$ — you cannot obtain $r$ successes in fewer than $r$ trials, so the support starts exactly at $r$ (matching $\left(\genfrac{}{}{0ex}{}{r-1}{r-1}\right)=1$).

A point mass at $k=r$: every trial succeeds, so the $r$-th success lands on trial $r$ exactly; mean $r/p\to r$ and variance $rq/p^2\to 0$.

Trivially, you need zero successes, so you stop before any trial: $X=0$ with certainty. It carries no information but confirms the "$r$ independent Geometric waits" picture — zero waits means zero trials.

A point mass at $X=1$: success is nearly certain immediately, so $P(X=1)=p\to 1$ and the mean $1/p\to 1$.

Yes — over the larger window both mean and variance become $2\lambda$, still equal; scaling the interval scales $\lambda$ but keeps the Poisson identity intact.