Binomial Models For Success Counts

From Binary Responses to a Success Count

A binary response records one of two outcomes, such as whether a patient improved or did not improve. In many studies, however, the question concerns several such responses together: how many successes occurred among a fixed number of trials? The count has a different support and a different probability model from any one binary response.

For example, if four patients each either respond to treatment or do not, each patient’s response is binary. The number who respond can be 0, 1, 2, 3, or 4. A binomial model describes this count when the trials satisfy specific conditions.

Definition: Let $Y_1,\ldots,Y_n$ be independent Bernoulli random variables with the same success probability $p$, where $0\leq p\leq 1$. The success count is $S=\sum_{i=1}^n Y_i$. It has a binomial distribution with size $n$ and success probability $p$, written $S\sim\operatorname{Bin}(n,p)$, if its probability mass function is $$ P(S=k)=\binom{n}{k}p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n. $$ Here $n$ is a fixed positive integer, and $\binom{n}{k}=n!/[k!(n-k)!]$.

The conditions in the definition matter. There must be a fixed number $n$ of trials; each trial must have the same probability $p$ of success; and the outcomes of the trials must be independent. The variable $Y_i$ describes the outcome of a single trial, taking the value 1 for success and 0 for failure. The variable $S$ describes the total number of successes across all $n$ trials.

Deriving the Binomial Probability Mass Function

To find the probability of exactly $k$ successes, consider the possible sequences of $n$ binary outcomes. Any particular sequence containing $k$ successes and $n-k$ failures has probability $p^k(1-p)^{n-k}$: independence lets us multiply the individual outcome probabilities. The event $S=k$ consists of all such sequences.

There are $\binom{n}{k}$ ways to choose which $k$ of the $n$ trial positions are successes. These sequences are mutually exclusive, and each has the same probability. Adding their probabilities gives the mass function.

Derivation: For any specified set of $k$ success positions, independence gives probability $p^k(1-p)^{n-k}$. There are $\binom{n}{k}$ choices of those positions. Since the choices give disjoint outcome sequences, $$ P(S=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$

The factor $\binom{n}{k}$ is essential: it accounts for how many different sequences produce the same count. For instance, with three trials, exactly two successes can occur as $110$, $101$, or $011$. Each sequence has the same probability, but the count event includes all three.

Worked Example: Exactly Two Responses Among Four Patients

Suppose four independent patients each have probability $p=0.3$ of responding to a treatment. Let $S$ be the number who respond. Find the probability that exactly two respond.

There are $n=4$ trials, all with the same success probability $0.3$, and the count of interest is $k=2$. The binomial formula gives $$ P(S=2)=\binom{4}{2}(0.3)^2(0.7)^2. $$

The number of sequences with two successes is $\binom{4}{2}=6$, and the probability of each such sequence is $(0.3)^2(0.7)^2=0.09(0.49)=0.0441$. Therefore $$ P(S=2)=6(0.0441)=0.2646. $$

The six sequences are $1100,1010,1001,0110,0101,0011$. Each represents a different assignment of response status to the four patients, but all contribute to the same count $S=2$.

Support and Probability Checks

The support of $S$ is $\{0,1,\ldots,n\}$: a count cannot be negative or exceed the number of trials. For $p=0$, the count is always 0; for $p=1$, it is always $n$. When $0<p<1$, every integer from 0 through $n$ has positive probability.

A probability mass function must sum to one over its support. For the binomial model, this follows directly from the binomial theorem.

Result: For $S\sim\operatorname{Bin}(n,p)$, the probabilities over all possible counts sum to one: $$ \sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}=1. $$ Proof: Apply the binomial theorem, $(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^k b^{n-k}$, with $a=p$ and $b=1-p$. Then $$ \sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k} =(p+(1-p))^n=1^n=1. $$ Thus the mass function assigns total probability one to its support. $\square$

Worked Example: At Least One Success

A screening test is positive for a particular condition with probability $0.2$ for each of three independent people. Let $S$ be the number of positive results. Find $P(S\geq 1)$.

The event $S\geq 1$ is the complement of $S=0$, so $$ P(S\geq 1)=1-P(S=0). $$

For zero positive results, all three tests must be negative. Each is negative with probability $1-0.2=0.8$, and independence gives $$ P(S=0)=\binom{3}{0}(0.2)^0(0.8)^3=1(1)(0.512)=0.512. $$

Therefore $$ P(S\geq 1)=1-0.512=0.488. $$ As a check, summing the probabilities for one, two, and three positives gives $3(0.2)(0.8)^2+3(0.2)^2(0.8)+(0.2)^3=0.384+0.096+0.008=0.488$.

What the Binomial Count Does—and Does Not—Represent

The individual response $Y_i$ and the aggregate count $S$ are different random variables. Each $Y_i$ has support $\{0,1\}$ and follows a Bernoulli model with success probability $p$. The count $S$ has support $\{0,1,\ldots,n\}$ and follows a binomial model under the stated independence and common-probability conditions. The count tells how many successes occurred, not which trials succeeded.

Quantity	What it describes	Possible values	Model
$Y_i$	Outcome on one specified trial	0 or 1	Bernoulli with probability $p$
$S=\sum_{i=1}^nY_i$	Total successes across $n$ trials	0 through $n$	Binomial with size $n$ and probability $p$

Worked Example: Individual Outcomes Versus Their Count

Three independent observations each have success probability $p=0.5$. Compare the event that the first observation is a success with the event that exactly one of the three observations is a success.

Let $Y_1$ record the first observation and let $S=Y_1+Y_2+Y_3$ be the total count. The individual response has $$ P(Y_1=1)=0.5. $$

For exactly one success in three trials, there are $\binom{3}{1}=3$ possible sequences, each with probability $(0.5)^1(0.5)^2=0.125$. Thus $$ P(S=1)=\binom{3}{1}(0.5)(0.5)^2=3(0.125)=0.375. $$

These probabilities differ because the events differ. $Y_1=1$ allows any outcomes on trials 2 and 3, whereas $S=1$ requires exactly one success in all three trials. The same data can be summarized either by the individual responses $(Y_1,Y_2,Y_3)$ or by their count $S$, but those summaries answer different questions.

Checking Whether a Binomial Model Fits

A count of successes is not automatically binomial. The model requires a fixed number of trials, a binary success-or-failure outcome for each trial, a common success probability, and independence. If the probabilities differ across trials, a particular sequence’s probability depends on which trials succeeded, so the simple factor $\binom{n}{k}$ times one common sequence probability no longer applies. If trials are dependent, multiplying their separate probabilities is not justified.

For example, suppose two patients have success probabilities $0.2$ and $0.8$, respectively, and their outcomes are independent. The probability that exactly one succeeds is $$ (0.2)(1-0.8)+(1-0.2)(0.8)=0.04+0.64=0.68. $$ A binomial model with $n=2$ would require one common probability $p$; it would give $2p(1-p)$, not the general expression above. Here no common $p$ describes the two trial probabilities.

Common mistake: Treating any total of binary outcomes as binomial. A success count has the binomial distribution only when the number of trials is fixed and the trials are independent with a common success probability. First check those conditions; only then use $\binom{n}{k}p^k(1-p)^{n-k}$.

Why the Distinction Matters

Choosing between a binary-response model and a count model starts with the response actually being analyzed. If each record concerns one trial, the response is binary. If one record summarizes a fixed group of trials by its number of successes, the response is a count. Confusing the two changes the support and assigns probabilities to the wrong events.

The binomial formula also clarifies why an aggregate count carries less detail than the full sequence. All sequences with $k$ successes are grouped into one event, and the combinatorial coefficient records how many sequences were grouped. Keeping both the count interpretation and the assumptions in view makes binomial probability calculations precise.

Check Your Understanding

Use the definition and assumptions to answer each question.

Four independent trials each have success probability $0.25$. Write the probability that exactly three succeed, leaving the answer in exact form.
For $n=5$, what values can the binomial count take? What values can one Bernoulli response take?
Explain why the event “the first trial succeeds” is not generally the same event as “exactly one success occurs among four trials.”
State the assumptions needed for a success count to have a $\operatorname{Bin}(n,p)$ distribution.
Two independent trials have success probabilities $0.1$ and $0.7$. Explain which binomial assumption fails, even though the total number of successes is a count.

Quantity	What it describes	Possible values	Model
\(Y_i\)	Outcome on one specified trial	0 or 1	Bernoulli with probability \(p\)
\(S=\sum_{i=1}^nY_i\)	Total successes across \(n\) trials	0 through \(n\)	Binomial with size \(n\) and probability \(p\)