Degenerate Binomial Cases and Bernoulli Equivalence

Why the Boundary Cases Matter

The binomial model is often introduced for a count that can take several values, from zero successes through $n$ successes. At the parameter boundaries, however, that count has no uncertainty: if success has probability zero, there are no successes; if success has probability one, every trial is a success. Recognizing these cases prevents unnecessary calculations and makes clear what the model says.

A second useful boundary is the trial count $n=1$. The count can then only be zero or one, so its distribution is the Bernoulli distribution. This equivalence lets a one-trial event be represented either as a binomial count or as a zero-one indicator.

Definition: For a positive integer $n$ and $0\leq p\leq 1$, $X\sim\operatorname{Bin}(n,p)$ denotes the number of successes in $n$ independent trials, each having the same success probability $p$. For $0<p<1$, its probability mass function is

P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n.

At $p=0$ or $p=1$, the distribution is defined by the trial outcomes directly; this avoids ambiguity about expressions such as $0^0$.

The binomial assumptions still matter at the endpoints: the variable counts successes across a fixed number of trials, and each trial has the same success probability. When $p=0$ or $p=1$, the outcomes are determined rather than random. In particular, the model assigns probability one to one count and probability zero to every other count.

Degenerate Binomial Distributions

A distribution concentrated at a single value is called degenerate. A binomial variable is degenerate at either probability boundary. The endpoint distributions, including their moments, are summarized below.

Parameters	Value of $X$ with probability one	$E[X]$	$\operatorname{Var}(X)$
$X\sim\operatorname{Bin}(n,0)$	0	0	0
$X\sim\operatorname{Bin}(n,1)$	$n$	$n$	0

These expectations and variances agree with the standard binomial formulas $E[X]=np$ and $\operatorname{Var}(X)=np(1-p)$. The direct description is especially useful for probabilities: with $p=0$, $P(X=0)=1$; with $p=1$, $P(X=n)=1$. Every other possible count has probability zero.

Theorem: If $X\sim\operatorname{Bin}(n,0)$, then $P(X=0)=1$. If $X\sim\operatorname{Bin}(n,1)$, then $P(X=n)=1$. In either case, $\operatorname{Var}(X)=0$, because a constant random variable has no variation.

Worked Example: A Success Probability of Zero

In a hypothetical quality-control exercise, six components are checked under a rule that makes the designated outcome “passes the special test” impossible. Let $X$ be the number that pass. Assume the checks are modeled as independent trials with the same success probability $p=0$. Thus $X\sim\operatorname{Bin}(6,0)$. Find $P(X=0)$, $P(X\geq 1)$, the mean, and the variance. No interest or mortality assumption applies.

Method. Use the degenerate endpoint distribution: with success probability zero, no trial can be a success.

Exact count. All six outcomes are failures, so $X=0$ with probability one. Therefore $P(X=0)=1$.

At least one success. The event $X\geq1$ cannot occur, so $P(X\geq1)=0$.

Mean and variance. Using the binomial formulas gives $E[X]=np=6(0)=0$ and $\operatorname{Var}(X)=np(1-p)=6(0)(1)=0$. These agree with the fact that $X$ is always zero.

Check. The two disjoint possibilities $X=0$ and $X\geq1$ have probabilities $1+0=1$. The common mistake avoided here is treating the endpoint as an ordinary uncertain case and assigning positive probability to a success.

Worked Example: A Success Probability of One

In a hypothetical training exercise, five records are reviewed, and each record is guaranteed to satisfy the designated criterion. Let $X$ count records satisfying it. Assume the binomial model with $n=5$ and $p=1$, so $X\sim\operatorname{Bin}(5,1)$. Find $P(X=5)$, $P(X\leq4)$, $E[X]$, and $\operatorname{Var}(X)$. No interest or mortality assumption applies.

Method. At the upper probability boundary, every trial is a success. Hence $X=5$ with probability one.

It follows that $P(X=5)=1$. Since the count cannot be four or less, $P(X\leq4)=0$.

The mean is $E[X]=np=5(1)=5$. The variance is $\operatorname{Var}(X)=np(1-p)=5(1)(0)=0$.

Check. The events $X=5$ and $X\leq4$ partition all possible counts, and their probabilities sum to $1+0=1$. The mistake to avoid is interpreting $p=1$ as a large but uncertain chance; it means success is certain on every trial under the stated model.

Why the Endpoint Results Hold

The endpoint cases can be justified from the trial interpretation without relying on any special convention for powers. If $p=0$, every trial is a failure, so the count is exactly zero. If $p=1$, every trial is a success, so the count is exactly $n$. A random variable that equals a constant $c$ with probability one has expectation $c$, and its variance is $\operatorname{Var}(X)=E[(X-c)^2]=E[0]=0$.

The probability mass function also gives the same result away from the potentially ambiguous $0^0$ term. For $p=0$ and $k>0$, the factor $p^k$ is zero, so those counts have probability zero. For $p=1$ and $k<n$, the factor $(1-p)^{n-k}$ is zero, so those counts have probability zero. The remaining endpoint count has probability one by the trial description.

Identify the endpoint.
Check whether the model has $p=0$ or $p=1$, rather than a probability strictly between zero and one.

Find the certain count.
For $p=0$, the certain count is 0; for $p=1$, it is $n$.

Evaluate events and moments.
An event containing the certain count has probability one; an event excluding it has probability zero. The mean is the certain count and the variance is zero.

One Trial: The Bernoulli Equivalence

A Bernoulli random variable records the outcome of one trial using the values zero and one. The value one represents success and zero represents failure. Its parameter $p$ is the probability of success, with $0\leq p\leq1$. When the binomial trial count is one, the binomial count uses exactly this same coding and has exactly this same distribution.

Definition: A Bernoulli random variable $B$ with success probability $p$, written $B\sim\operatorname{Bernoulli}(p)$, has

P(B=1)=p,\qquad P(B=0)=1-p.

For $X\sim\operatorname{Bin}(1,p)$, the possible counts are 0 and 1, and $X\sim\operatorname{Bernoulli}(p)$.

To verify the equivalence, put $n=1$ in the binomial probability formula. At $k=0$, $\binom{1}{0}=1$, giving probability $1-p$. At $k=1$, $\binom{1}{1}=1$, giving probability $p$. There are no other possible values. Thus the probability masses match the Bernoulli definition exactly.

P(X=0)=\binom{1}{0}p^0(1-p)^1=1-p,\qquad P(X=1)=\binom{1}{1}p^1(1-p)^0=p.

For $B\sim\operatorname{Bernoulli}(p)$, the value of $B$ itself is an indicator of success. Its mean is $E[B]=0(1-p)+1(p)=p$. Its second moment is $E[B^2]=0^2(1-p)+1^2p=p$, so $\operatorname{Var}(B)=E[B^2]-(E[B])^2=p-p^2=p(1-p)$. These agree with the binomial mean and variance for $n=1$.

The equivalence includes the endpoints: a Bernoulli variable with $p=0$ is always zero, and one with $p=1$ is always one. For $0<p<1$, it is not degenerate; both outcomes have positive probability. A one-trial count is therefore still a valid binomial model, even though Bernoulli notation may be more concise.

Worked Example: One Review and a Conditional Payment

A hypothetical administrator reviews one application. The probability that it meets a specified criterion is $0.30$. Let $X$ be the number of reviewed applications meeting the criterion. If the criterion is met, a fixed payment of $400 is made; otherwise, no payment is made. Assume the one review is a Bernoulli trial with the stated success probability, so $X\sim\operatorname{Bin}(1,0.30)$. Find the probability of each count, the probability of at least one success, and the expected payment. No interest or mortality assumption applies.

Method. Since $n=1$, identify $X$ as a Bernoulli variable with $p=0.30$. The count has values zero and one.

The exact probabilities are $P(X=1)=0.30$ and $P(X=0)=1-0.30=0.70$. With only one trial, “at least one success” means $X=1$, so $P(X\geq1)=0.30$.

Let $Y$ be the payment. It is $400 when $X=1$ and $0 when $X=0$, so $Y=400X$. Its expected value is $E[Y]=400E[X]=400(0.30)=\$120$.

Check. The two count probabilities sum to $0.70+0.30=1.00$, and the payment lies between $0 and $400, so an expected payment of $120 is consistent with the model. The common mistake avoided is treating “at least one” as a multi-trial tail event: with one trial, it is simply the success event.

Common Mistakes and Exam Traps

Boundary values and the one-trial case are simple, but a few notation errors can lead to incorrect answers.

Substituting blindly into the mass function. At $p=0,k=0$, direct substitution includes $0^0$. Declaring $0^0=0$ would incorrectly give $P(X=0)=0$. Use the trial interpretation: when $p=0$, every trial fails, so $P(X=0)=1$.
Confusing the boundary $p=1$ with $n=1$. The condition $p=1$ makes a binomial count degenerate. The condition $n=1$ makes it Bernoulli, which can still have uncertainty whenever $0<p<1$.
Assigning probability to impossible counts. For $X\sim\operatorname{Bin}(n,0)$, counts above zero are impossible. For $X\sim\operatorname{Bin}(n,1)$, counts below $n$ are impossible.
Forgetting the event definition. “At least one” means $X\geq1$; when $n=1$, this is the same event as $X=1$. The phrasing does not create additional possible outcomes.
Changing the meaning of success without changing the variable. If success is recoded as failure, the zero-one variable changes. State clearly which outcome is coded as one before assigning its Bernoulli parameter.

Key takeaway: For $X\sim\operatorname{Bin}(n,p)$, the endpoint cases are deterministic: $p=0$ gives $X=0$, and $p=1$ gives $X=n$, each with probability one. When $n=1$, the binomial count has the Bernoulli distribution with success probability $p$.

Check Your Understanding

For each problem, state the relevant model and use the endpoint or one-trial result where appropriate.

A hypothetical inspection program records $X\sim\operatorname{Bin}(4,0)$, the number of items meeting a criterion. Find $P(X=0)$, $P(X\geq1)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $1,\ 0,\ 0,\ 0$, respectively.)
In a hypothetical audit, $X\sim\operatorname{Bin}(7,1)$. Find $P(X=7)$, $P(X\leq6)$, and the mean and variance. (Answer: $1,\ 0,\ E[X]=7,\ \operatorname{Var}(X)=0$.)
A single hypothetical trial has success probability $0.42$, and $X\sim\operatorname{Bin}(1,0.42)$. Give its Bernoulli parameter, $P(X=0)$, $P(X=1)$, and $P(X\geq1)$. (Answer: $\operatorname{Bernoulli}(0.42),\ 0.58,\ 0.42,\ 0.42$.)
One hypothetical review succeeds with probability $0.25$. A payment $Y$ is $120 if it succeeds and $0 otherwise. Find $E[Y]$, $P(Y\geq100)$, and $\operatorname{Var}(Y)$. (Answer: $E[Y]=\$30,\ P(Y\geq100)=0.25,\ \operatorname{Var}(Y)=\$2{,}700$ dollars squared.)
A single hypothetical test has probability $0.64$ of producing no alert. Let $X=1$ denote “no alert” and $W=1-X$ denote “alert.” Give the Bernoulli parameter of $W$, then find $P(W=0)$, $P(W=1)$, and $\operatorname{Var}(W)$. (Answer: $W\sim\operatorname{Bernoulli}(0.36)$, $P(W=0)=0.64$, $P(W=1)=0.36$, and $\operatorname{Var}(W)=0.36(0.64)=0.2304$.)

Parameters	Value of \(X\) with probability one	\(E[X]\)	\(\operatorname{Var}(X)\)
\(X\sim\operatorname{Bin}(n,0)\)	0	0	0
\(X\sim\operatorname{Bin}(n,1)\)	\(n\)	\(n\)	0

What You'll Learn