Using the Binomial Cumulative Distribution

Why Cumulative Probabilities Are Useful

A binomial probability can ask for one exact count, such as exactly three successes, or for a collection of counts, such as no more than three or between two and five. For a range, adding every exact probability is valid, but it can be lengthy. The cumulative distribution function (CDF) packages the probabilities up to a cutoff into one quantity. Subtracting two cumulative probabilities then isolates the counts in an interval.

The method applies when the count $X$ follows a binomial model: there are a fixed number $n$ of trials, each trial has two outcomes, the trials are independent, and the probability of success $p$ is the same on every trial. Thus $X\sim\operatorname{Bin}(n,p)$, with possible values $0,1,\ldots,n$. The exact probability at count $k$ is $\binom{n}{k}p^k(1-p)^{n-k}$. The cumulative method adds these exact probabilities over the counts required by the event.

This tutorial focuses on calculating cumulative and interval probabilities, not on selecting a cutoff to meet a target probability. The methods are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition and Interval Rules

Definition: For a random variable $X$, its cumulative distribution function is $F_X(x)=P(X\le x)$. For $X\sim\operatorname{Bin}(n,p)$ and an integer $k$ with $0\le k\le n$, $$ F_X(k)=P(X\le k)=\sum_{j=0}^{k}\binom{n}{j}p^j(1-p)^{n-j}. $$ For $k<0$, $F_X(k)=0$; for $k\ge n$, $F_X(k)=1$.

Because a binomial count only takes integer values, the CDF stays constant between consecutive integers. For integer endpoints $a\le b$, the probability of an inclusive interval is

P(a\le X\le b)=F_X(b)-F_X(a-1).

The subtraction removes every count below $a$, including $a-1$, while retaining the probability at $a$. For a strict lower bound and an inclusive upper bound,

P(a<X\le b)=F_X(b)-F_X(a).

More generally, if the stated endpoints are not integers, use the integer counts that actually satisfy the event. For example, $2.2\le X\le 5.7$ is the same event as $3\le X\le5$, since $X$ cannot take noninteger values. One can therefore use $F_X(5)-F_X(2)$.

How Subtraction Isolates a Range

The interval rule follows by separating the event $X\le b$ into two disjoint parts: $X\le a-1$ and $a\le X\le b$. Their probabilities add, so subtracting the first part's probability leaves exactly the interval probability.

\begin{aligned} \{X\le b\}&=\{X\le a-1\}\ \cup\ \{a\le X\le b\},\\ P(X\le b)&=P(X\le a-1)+P(a\le X\le b),\\ F_X(b)&=F_X(a-1)+P(a\le X\le b),\\ P(a\le X\le b)&=F_X(b)-F_X(a-1). \end{aligned}

The union is disjoint: an integer count cannot be both at most $a-1$ and at least $a$. This justification also explains the most important endpoint detail. To include $a$, subtract only through $a-1$, not through $a$.

Worked Cumulative and Interval Calculations

Worked Example: A Short Cumulative Calculation

A hypothetical quality check inspects five independent components. Each component has probability $0.2$ of failing, independently of the others. Let $X$ be the number that fail. Find the probability that between one and three components fail, inclusive.

Model and method. The number of trials is $n=5$, the failure probability is $p=0.2$, and $X\sim\operatorname{Bin}(5,0.2)$. The event is $1\le X\le3$, so use $F_X(3)-F_X(0)$.

First calculate the exact probabilities needed for the cumulative values. Here $1-p=0.8$:

\begin{aligned} P(X=0)&=\binom{5}{0}(0.2)^0(0.8)^5=0.32768,\\ P(X=1)&=\binom{5}{1}(0.2)(0.8)^4=5(0.2)(0.4096)=0.4096,\\ P(X=2)&=\binom{5}{2}(0.2)^2(0.8)^3=10(0.04)(0.512)=0.2048,\\ P(X=3)&=\binom{5}{3}(0.2)^3(0.8)^2=10(0.008)(0.64)=0.0512. \end{aligned}

Therefore $F_X(0)=0.32768$, and

$$ F_X(3)=0.32768+0.4096+0.2048+0.0512=0.99328. $$

Subtract the cumulative probability below the lower endpoint:

P(1\le X\le3)=F_X(3)-F_X(0)=0.99328-0.32768=0.66560.

Rounded only at the final step, the probability is 0.6656 to four decimal places. As a check, adding the probabilities at counts 1, 2, and 3 gives $0.4096+0.2048+0.0512=0.6656$. The cumulative method avoids adding every required count separately. The common mistake to avoid is subtracting $F_X(1)$, which would remove the count 1 that the event includes.

Worked Example: A Range With Several Possible Counts

A hypothetical review examines eight independent applications. Each application has probability $0.25$ of requiring an additional document. Let $X$ count applications requiring that document. Find the probability that from two through five applications, inclusive, require it.

Model and method. Here $n=8$, $p=0.25$, and $X\sim\operatorname{Bin}(8,0.25)$. The event is $2\le X\le5$, so calculate $F_X(5)-F_X(1)$. To keep the arithmetic exact, write each probability with denominator $4^8=65{,}536$, since $p=1/4$ and $1-p=3/4$.

For each count $k$, the numerator is $\binom{8}{k}3^{8-k}$. The values from zero through five are:

\begin{aligned} P(X=0)&=\frac{3^8}{65{,}536}=\frac{6{,}561}{65{,}536},& P(X=1)&=\frac{8(3^7)}{65{,}536}=\frac{17{,}496}{65{,}536},\\ P(X=2)&=\frac{28(3^6)}{65{,}536}=\frac{20{,}412}{65{,}536},& P(X=3)&=\frac{56(3^5)}{65{,}536}=\frac{13{,}608}{65{,}536},\\ P(X=4)&=\frac{70(3^4)}{65{,}536}=\frac{5{,}670}{65{,}536},& P(X=5)&=\frac{56(3^3)}{65{,}536}=\frac{1{,}512}{65{,}536}. \end{aligned}

The cumulative probability through one is

F_X(1)=\frac{6{,}561+17{,}496}{65{,}536} =\frac{24{,}057}{65{,}536}.

The cumulative probability through five is

F_X(5)=\frac{6{,}561+17{,}496+20{,}412+13{,}608+5{,}670+1{,}512}{65{,}536} =\frac{65{,}259}{65{,}536}.

Now subtract:

P(2\le X\le5)=F_X(5)-F_X(1) =\frac{65{,}259-24{,}057}{65{,}536} =\frac{41{,}202}{65{,}536} \approx 0.6287.

The final probability is 0.6287 to four decimal places. As a check, the four numerators for counts 2 through 5 add to $20{,}412+13{,}608+5{,}670+1{,}512=41{,}202$, the same interval numerator. The method avoids the common endpoint error of subtracting $F_X(2)$, which would wrongly discard count 2.

Worked Example: Interpreting Strict and Inclusive Bounds

A hypothetical screening program checks nine independent records, each of which has probability $0.5$ of containing a specified flag. Let $X$ be the number of flagged records. Find the probability that at least three but no more than seven records are flagged.

Model and method. The count is $X\sim\operatorname{Bin}(9,0.5)$. “At least three” means $X\ge3$, while “no more than seven” means $X\le7$; together they give $3\le X\le7$. Use $F_X(7)-F_X(2)$.

For $p=0.5$, each count probability is its binomial coefficient divided by $2^9=512$. The coefficients for counts zero through nine are $1,9,36,84,126,126,84,36,9,1$. Thus,

F_X(2)=\frac{1+9+36}{512}=\frac{46}{512}.

Through count seven, it is efficient to subtract the last two probabilities from the total probability 1:

F_X(7)=1-\frac{9+1}{512}=\frac{502}{512}.

Therefore,

P(3\le X\le7)=F_X(7)-F_X(2) =\frac{502-46}{512} =\frac{456}{512} =0.890625.

Rounded only at the final step, the answer is 0.8906 to four decimal places. A direct check adds the coefficients for counts 3 through 7: $84+126+126+84+36=456$. A common mistake is to read “at least three” as “greater than three”; the event includes count 3, so the subtracted CDF must stop at 2.

Practical Checks and Common Traps

A CDF calculation is easiest to audit when the event is translated into integer counts before any probabilities are evaluated. Use this sequence:

Confirm the binomial model.
Identify the fixed trial count $n$, common success probability $p$, and the count $X$. The binomial CDF is not appropriate if the stated binomial assumptions do not hold.

Translate the event.
Rewrite verbal bounds as integer values of $X$. Keep track of whether “at least,” “more than,” “at most,” or “fewer than” includes the endpoint.

Choose the CDF subtraction.
For $a\le X\le b$, calculate $F_X(b)-F_X(a-1)$. For $a<X\le b$, calculate $F_X(b)-F_X(a)$.

Check the result.
Confirm the probability is between 0 and 1. When practical, compare it with a sum of the exact probabilities in the requested range.

One frequent error is using the right CDF values with the wrong lower endpoint. For the second example, an incorrect calculation for $2\le X\le5$ would be

F_X(5)-F_X(2) =\frac{65{,}259-44{,}469}{65{,}536} =\frac{20{,}790}{65{,}536} \approx 0.3172.

This subtracts the probability at count 2 as well as all lower counts, so it answers a different question. The correction is to subtract through count $2-1=1$:

F_X(5)-F_X(1) =\frac{65{,}259-24{,}057}{65{,}536} \approx 0.6287.

The subtraction rule is a direct consequence of disjoint events, not a special calculator convention. If a calculator or software package provides a binomial cumulative function, check whether it returns $P(X\le k)$; the notation and input conventions should be verified before using its output.

Key takeaway: For an inclusive integer range $a\le X\le b$, subtract $F_X(a-1)$ from $F_X(b)$. Translating the wording into integer endpoints first prevents most cumulative-probability errors.

Check Your Understanding

For each problem, define the count as binomial under the stated independent, equal-probability trial assumptions. Round probability answers to four decimal places.

A hypothetical device has six independent components, each with probability $0.5$ of needing adjustment. Find the probability that between two and four components, inclusive, need adjustment. (Answer: 0.7813)
In a hypothetical group of five independent records, each has probability $0.2$ of a coding error. Find the probability of at most two coding errors. (Answer: 0.9421)
Eight independent applications each have probability $0.25$ of a missing attachment. Find the probability that more than one but no more than five applications have a missing attachment. (Answer: 0.6287)
Six independent inspections each have probability $0.4$ of detecting a defect. Find the probability that at least two but fewer than six inspections detect one. (Answer: 0.7626)
Four independent test units each have probability $0.3$ of triggering an alert. Find the probability that more than zero but fewer than three units trigger an alert. (Answer: 0.6762)

What You'll Learn