Common distributions (Bernoulli, Binomial, Poisson): Question bank

This page is a concept gym for Common distributions (Bernoulli, Binomial, Poisson). Every item below is a question ::: answer reveal — read the left side, think, then check. None of these need a calculator; they attack the ideas, the assumptions, and the boundary cases where beginners slip.

Before we start, let's earn every symbol these questions will use. Nothing below appears in a question until it has a plain-word meaning and a legal set of values.

Definition The symbols and their legal values (domains)

$p$ — the probability of success on a single trial. Being a probability, it must satisfy $0 \leq p \leq 1$ .
$n$ — the number of trials, always a positive whole number: $n \in {1, 2, 3, \dots}$ (written $n \in N$ ).
$k$ — a specific count we ask about, e.g. "exactly $k$ successes." For a Binomial it must be a whole number in ${0, 1, \dots, n}$ ; for a Poisson any whole number $k \in {0, 1, 2, \dots}$ (written $N_{0}$ , the naturals including zero). Because $k$ counts things, fractions are illegal.
Rate $λ$ ("lambda") — the average count of events expected in one fixed chunk of time or space. It is an average of counts, so it can be zero but never negative: $λ \geq 0$ . " $3$ errors per hour" means $λ = 3$ per one-hour interval.
Mean $E [X]$ — the long-run average value of the count if you repeated the experiment forever.
Variance $Var (X)$ — how far, on average, the count spreads from its mean; its square root is the standard deviation.

Now all three PMFs (probability mass functions), stated with their domains and the one-line reason each factor exists — so the intuition sits right next to the symbol:

The figure below draws all three shapes on one axis so you can see the domains: Bernoulli lives only on ${0, 1}$ (two open squares), Binomial fills ${0, \dots, n}$ and then stops hard at $k = n$ , while Poisson (red) marches off to the right with no cap. Keep this picture in mind — several questions below hinge directly on it.

Figure — Common distributions (Bernoulli, Binomial, Poisson)

True or false — justify

A Binomial random variable can never be larger than

n

True. You count successes among exactly

n

trials, so the count lives in

{0, 1, \dots, n}

— it is capped at

n

. In the figure, the black Binomial stems end abruptly at

k = n = 10

A Poisson random variable has a largest possible value

False. Poisson allows

k = 0, 1, 2, \dots

with no upper limit; every count has some (shrinking) positive probability. The red stems in the figure keep going rightward — there is no "maximum number of errors."

A Bernoulli variable is just a Binomial with

n = 1

True. One trial, count of successes = 0 or 1. Plugging

n = 1

into

E [X] = n p

and

Var (X) = n p (1 - p)

recovers the Bernoulli mean

p

and variance

p (1 - p)

For a fair coin, the variance of a single flip is

0.25

True.

p = 0.5

gives

p (1 - p) = 0.5 \times 0.5 = 0.25

, the maximum possible Bernoulli variance — maximum uncertainty at 50/50.

If mean equals variance, the data must be Poisson

False. Poisson forces

E [X] = Var (X) = λ

, but the reverse isn't guaranteed — other distributions can accidentally match. It's a necessary clue, not a proof.

Increasing

n

in a Binomial always increases the variance

False as stated — it needs the qualifier $0 < p < 1$ . For those

p

, variance

n p (1 - p)

does grow linearly with

n

. But at the boundaries

p = 0

p = 1

the factor

p (1 - p) = 0

, so

Var = 0

for every

n

— the count is fixed and never wobbles.

The Poisson PMF sums to 1 over all

k \geq 0

True.

\sum_{k} \frac{λ ^{k} e ^{- λ}}{k !} = e^{- λ} \sum_{k} \frac{λ ^{k}}{k !} = e^{- λ} e^{λ} = 1

, using the Taylor series for

e^{λ}

A Binomial with

p = 0

is a valid distribution

True but degenerate. Every trial fails, so

P (X = 0) = 1

and

X

is always

0

. Mean

= n p = 0

, variance

= n p (1 - p) = 0

: no randomness left.

Doubling

λ

doubles both the mean and the variance of a Poisson

True. Both equal

λ

, so scaling

λ

scales both identically. This is exactly the "mean

=

variance" fingerprint at work.

Spot the error

P (X = k) = (k n) p^{k} (1 - p)^{n - k}

, so

P (X = 1.5)

is fine." — what's wrong?

X

is a count of successes, so

k

must be a whole number in

{0, \dots, n}

. There is no "1.5 successes"; the PMF is only defined on integers.

"For rare disease modelling I'll use Binomial with

n = 1, 000, 000

." — what's the issue?

Not wrong, but the huge

n

with tiny

p

makes the binomial coefficient computationally brutal. Since

n p = λ

is finite and

p

is tiny, Poisson gives essentially the same answer far more cheaply.

"I flip a coin 10 times but chase losses (gambler's fallacy), still Binomial." — flaw?

Binomial requires independence. If your belief about trial

i

depends on earlier results, the independence assumption is violated and the count is no longer Binomial.

"Poisson mean is

λ

, so variance must be

λ^{2}

." — where's the slip?

For Poisson the variance is

λ

, not

λ^{2}

. Mean and variance are equal, which is the distribution's signature — don't confuse it with squaring the mean.

P (X = x) = p^{x} (1 - p)^{1 - x}

works for any

x

." — the catch?

This Bernoulli PMF is only meaningful for

x \in {0, 1}

; the exponents are a trick to select

p

1 - p

. Plug in

x = 2

and you get a meaningless number, not a probability.

"Binomial standard deviation is

n p (1 - p)

." — mistake?

That's the variance. The standard deviation is its square root,

n p (1 - p)

— deviations are measured in the same units as the count, not squared units.

"For

λ = 3

P (X = 0) = 0

because zero errors is impossible." — error?

P (X = 0) = e^{- λ} = e^{- 3} \approx 0.05

, which is small but positive. A perfectly clean hour is unlikely, not impossible.

Why questions

Why does Bernoulli variance

p (1 - p)

peak at

p = 0.5

That's the point of maximum uncertainty — both outcomes equally likely, so you can least predict the result. Near

p = 0

p = 1

the outcome is nearly certain, so there's little to be surprised by.

Why is Binomial the sum of Bernoullis?

Counting successes over

n

trials is literally adding up

n

individual 0/1 indicators

X_{1} + \dots + X_{n}

. This is why linearity of expectation gives mean

= n p

instantly.

Why does the

(1 - \frac{λ}{n})^{n}

factor become

e^{- λ}

in the Poisson limit?

It matches the classic limit

(1 + \frac{x}{n})^{n} \to e^{x}

with

x = - λ

. This is the very definition of

e

, and it's why the exponential appears in Poisson at all.

Why can't we add variances of trials unless they're independent?

The rule

Var (X_{1} + \dots + X_{n}) = \sum Var (X_{i})

only holds when trials share no correlation. If they influence each other, cross-covariance terms appear and the neat

n p (1 - p)

breaks.

Why does Poisson need "events happen at a constant rate"?

The single parameter

λ

encodes one fixed average per interval. If the rate surges (e.g. traffic spikes at noon), one

λ

can't describe both quiet and busy periods, and the model misfits.

Why is

(k n)

needed at all in the Binomial PMF?

p^{k} (1 - p)^{n - k}

is the probability of one specific sequence of

k

successes. There are many orderings that give

k

successes, and

(k n)

counts them so we add up every arrangement.

Why does Poisson replace

n

and

p

with a single

λ

In the rare-event limit only the product

n p = λ

(expected count) survives — the individual values of

n

and

p

stop mattering. So we describe everything with one number.

Edge cases

What is

(0 n)

and why?

It equals

1

— there is exactly one way to choose zero successes (choose nothing). Using

(0 n) = \frac{n !}{0 ! n !} = 1

makes

P (X = 0) = (1 - p)^{n}

come out correctly for the Binomial.

What does

0!

equal, and where does it bite?

0! = 1

by convention (the empty product), which keeps

P (X = 0) = \frac{λ ^{0} e ^{- λ}}{0 !} = e^{- λ}

well-defined for Poisson. Treating it as

0

would wrongly blow the probability up.

What happens to a Binomial when

p = 1

Every trial succeeds, so

X = n

with probability 1. Mean

= n

, variance

= n p (1 - p) = 0

: fully certain, zero spread.

Is a Poisson with

λ = 0

meaningful?

It degenerates:

P (X = 0) = e^{0} = 1

and all other counts have probability 0. A rate of zero means the event never occurs — consistent with

λ \geq 0

allowing the boundary value.

Can

p

be exactly

0.5

and still give a skewed Binomial?

No. At

p = 0.5

the Binomial is perfectly symmetric about

n /2

. Skew appears only when

p \neq = 0.5

, leaning toward the more likely outcome.

What is the smallest possible value a Poisson variable can take, and can it be negative?

The smallest value is

0

; it can never be negative because you cannot count a negative number of events. Counts start at zero and go up.

If you flip one coin, why is Binomial variance formula still consistent?

With

n = 1

n p (1 - p) = p (1 - p)

, which is exactly the Bernoulli variance. The formulas dovetail because a single flip is the base case.

Recall Rapid self-check

Poisson has an upper bound on its count. ::: False — counts run to infinity. Binomial variance is $n p (1 - p)$ . ::: False — that's the standard deviation; variance is $n p (1 - p)$ . $P (X = 0)$ for Poisson equals $e^{- λ}$ . ::: True. Mean equals variance proves data is Poisson. ::: False — necessary clue, not proof. Increasing $n$ always raises Binomial variance. ::: False — only for $0 < p < 1$ ; at $p = 0$ or $p = 1$ variance stays $0$ .