Verifying Binomial Probabilities With the Binomial Theorem

Why Check the Total Probability?

A probability mass function must assign probabilities that add to one across all possible outcomes. For a binomial count, those outcomes are the possible numbers of successes. Showing that their probabilities sum to one is more than an arithmetic check: it verifies that the formula accounts for every possible count without assigning probability to impossible counts.

The binomial theorem gives a short general proof. Its terms match the binomial probability formula, and its expansion turns the total into a familiar power. The same result can also be checked by listing the probabilities for a small number of trials. Those calculations make the coefficients and exponents visible, which helps catch common errors.

Model and Probability Formula

Suppose $n$ trials are performed, where $n$ is a nonnegative integer. Each trial has two outcomes, designated success and failure. Assume the trials are independent and that the probability of success is the same value $p$ on every trial, with $0\leq p\leq 1$. Let $X$ be the number of successes. Under these assumptions, $X$ has a binomial distribution with parameters $n$ and $p$.

Definition: For $X\sim\operatorname{Bin}(n,p)$, the possible values of $X$ are the integers $0,1,\ldots,n$, and its probability mass function is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. $$ Here, $\binom{n}{k}$ counts the ways to select which $k$ of the $n$ trials are successes. The formula does not apply as a binomial model if the trials are dependent or their success probabilities differ.

The probability of any one specified success pattern with $k$ successes is $p^k(1-p)^{n-k}$: independence lets us multiply the probability $p$ for each success and $1-p$ for each failure. There are $\binom{n}{k}$ such patterns, and they are mutually exclusive. Adding their probabilities gives the formula for $P(X=k)$.

Deriving the Normalization Identity

The binomial theorem states that for any numbers $a$ and $b$, and any nonnegative integer $n$,

(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^k b^{n-k}.

To sum the binomial probabilities, use $a=p$ and $b=1-p$. The resulting terms are exactly the probabilities for $k=0,1,\ldots,n$:

\begin{aligned} \sum_{k=0}^{n}P(X=k) &=\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}\\ &=(p+(1-p))^n\\ &=1^n\\ &=1. \end{aligned}

This argument relies on including every possible count, from zero successes through $n$ successes. These events are mutually exclusive, and together they cover all possible outcomes of the $n$ trials. The sum is therefore the total probability of the sample space, and the theorem evaluates that sum as one.

The result also holds at the parameter endpoints. If $p=0$, the count is always zero; if $p=1$, it is always $n$. In each case one possible count has probability one and all the others have probability zero. The identity remains valid, with the usual interpretation of the distribution at those endpoints.

Worked Checks by Expansion

Worked Example: Three Hypothetical Quality-Control Tests

A hypothetical inspection process independently tests three items. Each item has probability $0.2$ of being classified as defective. Let $X$ be the number classified as defective. Verify that the probabilities for all possible values of $X$ sum to one.

Method. The model is $X\sim\operatorname{Bin}(3,0.2)$. The trials are independent, have the same success probability, and $X$ can be $0,1,2,$ or $3$. Write the mass function at each value, then add the terms.

\begin{aligned} P(X=0)&=\binom{3}{0}(0.2)^0(0.8)^3=1(1)(0.512)=0.512,\\ P(X=1)&=\binom{3}{1}(0.2)^1(0.8)^2=3(0.2)(0.64)=0.384,\\ P(X=2)&=\binom{3}{2}(0.2)^2(0.8)^1=3(0.04)(0.8)=0.096,\\ P(X=3)&=\binom{3}{3}(0.2)^3(0.8)^0=1(0.008)(1)=0.008. \end{aligned}

Adding in order gives $0.512+0.384=0.896$, then $0.896+0.096=0.992$, and finally $0.992+0.008=1.000$. The probabilities sum to one, as the binomial theorem predicts. The check also confirms that the largest possible count, three, was included.

Common mistake avoided: forgetting the coefficient $\binom{3}{1}=3$ or $\binom{3}{2}=3$. Each probability for exactly one or exactly two defects includes several possible patterns, not just one specified sequence.

Worked Example: Four Hypothetical Sensor Readings

A hypothetical monitoring system records four independent readings. Each reading has probability $0.3$ of triggering an alert. Let $X$ be the number of alerts. Check the total probability by expanding the binomial theorem.

Method. Here $X\sim\operatorname{Bin}(4,0.3)$, and a non-alert has probability $1-0.3=0.7$. The possible counts run from zero through four. Compute each term, keeping the exponent of $0.3$ equal to the count and the exponent of $0.7$ equal to the remaining number of readings.

\begin{aligned} P(X=0)&=\binom{4}{0}(0.3)^0(0.7)^4=1(1)(0.2401)=0.2401,\\ P(X=1)&=\binom{4}{1}(0.3)^1(0.7)^3=4(0.3)(0.343)=0.4116,\\ P(X=2)&=\binom{4}{2}(0.3)^2(0.7)^2=6(0.09)(0.49)=0.2646,\\ P(X=3)&=\binom{4}{3}(0.3)^3(0.7)^1=4(0.027)(0.7)=0.0756,\\ P(X=4)&=\binom{4}{4}(0.3)^4(0.7)^0=1(0.0081)(1)=0.0081. \end{aligned}

The partial sums are $0.2401+0.4116=0.6517$, $0.6517+0.2646=0.9163$, $0.9163+0.0756=0.9919$, and $0.9919+0.0081=1.0000$. Thus the complete set of probabilities is normalized.

Common mistake avoided: using $(0.7)^4$ in every term. The failure probability is raised to $4-k$, not to four, because a count of $k$ alerts leaves exactly $4-k$ non-alerts.

Worked Example: A Symbolic Check for Five Trials

A hypothetical process consists of five independent trials, each with success probability $p=3/7$. Verify the total probability using both the general identity and an explicit expansion.

Method. The complement probability is $1-p=4/7$, and $X\sim\operatorname{Bin}(5,3/7)$. The binomial theorem gives the total immediately; the explicit terms provide an arithmetic check.

\begin{aligned} \sum_{k=0}^{5}P(X=k) &=\sum_{k=0}^{5}\binom{5}{k}\left(\frac{3}{7}\right)^k \left(\frac{4}{7}\right)^{5-k}\\ &=\left(\frac{3}{7}+\frac{4}{7}\right)^5=1. \end{aligned}

For the term-by-term check, the six numerators over the common denominator $7^5=16807$ are:

\begin{aligned} P(X=0)&=\frac{1024}{16807},& P(X=1)&=\frac{3840}{16807},& P(X=2)&=\frac{5760}{16807},\\ P(X=3)&=\frac{4320}{16807},& P(X=4)&=\frac{1620}{16807},& P(X=5)&=\frac{243}{16807}. \end{aligned}

The numerator sum is $1024+3840=4864$, $4864+5760=10624$, $10624+4320=14944$, $14944+1620=16564$, and $16564+243=16807$. Therefore the probabilities total $16807/16807=1$, agreeing with the theorem.

Common mistake avoided: omitting the endpoints $k=0$ or $k=5$. The endpoint terms correspond to all failures and all successes; both are part of the distribution even when one seems unlikely.

What Normalization Can and Cannot Confirm

For an exam calculation, the theorem offers a compact check: if the probabilities have the binomial form and all counts are included, their sum must be one. An expanded calculation is useful for detecting a specific arithmetic error, while the theorem proves the identity for every valid $n$ and $p$ at once.

However, a sum of one by itself does not establish that a proposed list is the correct binomial distribution. Incorrect terms can sometimes still add to one. Check both the model assumptions and each term's coefficient and exponents. In particular, confirm that:

The count $k$ ranges over every integer from $0$ to $n$.
The coefficient is $\binom{n}{k}$, accounting for all patterns with $k$ successes.
The success and failure factors are $p^k$ and $(1-p)^{n-k}$, respectively.
The success probability is common to the trials and the trials are independent.

The normalization identity concerns the total probability only. It does not, by itself, calculate a particular tail probability or summarize the distribution with an average or variance. Its role is foundational: it confirms that the binomial mass function allocates a complete unit of probability across its support.

Key takeaway: The binomial probabilities sum to one because their sum is the binomial expansion of $(p+(1-p))^n=1$. For small $n$, expanding each term is an effective way to check coefficients, exponents, and endpoints.

Check Your Understanding

For each problem, identify the full range of possible counts and verify or state the total probability.

A hypothetical device is tested twice independently, with probability $0.4$ of a fault on each test. Write the three binomial probabilities and check their sum. (Answer: $0.36+0.48+0.16=1.00$.)
Three independent hypothetical applications each have probability $0.25$ of being approved. Write the four probabilities for the number approved and verify their total. (Answer: $0.421875+0.421875+0.140625+0.015625=1.000000$.)
A hypothetical sensor produces four independent signals, each with probability $0.1$ of being positive. Give the probabilities for zero through four positive signals and check their sum. (Answer: $0.6561+0.2916+0.0486+0.0036+0.0001=1.0000$.)
For $X\sim\operatorname{Bin}(6,0.7)$, use the binomial theorem to evaluate $\sum_{k=0}^{6}P(X=k)$, and state which two quantities are substituted for $a$ and $b$. (Answer: $1$, with $a=0.7$ and $b=0.3$.)
A proposed formula for $X\sim\operatorname{Bin}(4,0.2)$ uses $\binom{4}{k}(0.2)^k(0.8)^4$ for every $k$. Identify the error and give the correct exponent on $0.8$. (Answer: The failure exponent is wrong; it must be $4-k$, so the correct factor is $(0.8)^{4-k}$.)

What You'll Learn