Cumulative Binomial Probabilities and Interval Events

From One Count to a Range of Counts

A binomial probability calculation can ask about one exact count, such as the probability of exactly three successes. Many practical questions concern a range instead: no more than three successes, at least six successes, or between two and five successes. These are cumulative binomial probabilities: probabilities found by combining the probabilities of several possible counts.

The key is to identify every integer count included in the event. A count is always an integer, so the wording of the question determines whether the endpoint counts are included. Once the included values are clear, add their individual probabilities.

Definition: If $X$ is a binomial random variable with $n$ trials and success probability $p$ on each trial, then its possible values are the integers $0,1,\ldots,n$. For any set $A$ of these values, $P(X\in A)$ is the sum of the point probabilities $P(X=k)$ over all $k$ in $A$. In particular, “at most $c$” means $X\le c$, “at least $c$” means $X\ge c$, and “between $a$ and $b$, inclusive” means $a\le X\le b$.

The Model and the Addition Rule

Before calculating, confirm that the binomial model describes the count. It applies when there is a fixed number $n$ of trials, each trial has two possible outcomes, the trials are independent, and the probability $p$ of success is the same on every trial. The random variable $X$ counts the successes. Under these conditions, $X\sim\operatorname{Bin}(n,p)$, and the point probability for an integer $k$ from 0 through $n$ is

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

This point-probability formula gives the probability of one particular count. To find the probability of a range, use the fact that the events $X=k$ for different values of $k$ cannot happen together. For example, a single set of trials cannot result in both exactly two successes and exactly three successes. These events are disjoint, so the probability of their union is the sum of their probabilities.

Result: For integers $L$ and $U$ with $0\le L\le U\le n$, $$ P(L\le X\le U)=\sum_{k=L}^{U}\binom{n}{k}p^k(1-p)^{n-k}. $$ The corresponding lower and upper cumulative probabilities are $P(X\le U)=\sum_{k=0}^{U}P(X=k)$ and $P(X\ge L)=\sum_{k=L}^{n}P(X=k)$.

Why the Probabilities Add

The addition used here follows from the basic rule for disjoint events. If events $A_1,\ldots,A_m$ are mutually exclusive, then $P(A_1\cup\cdots\cup A_m)=P(A_1)+\cdots+P(A_m)$. A range of binomial counts is exactly such a union: for example, the event $1\le X\le 3$ is the union of $X=1$, $X=2$, and $X=3$.

This justifies both the summation and its limits. The lower limit names the first included count; the upper limit names the last included count. Counts outside the possible range $0,\ldots,n$ are not possible outcomes of the model.

Calculating an “At Most” Probability

“At most $c$” includes $c$ itself and every smaller possible count. Thus, for an integer $c$ between 0 and $n$, calculate $P(X\le c)$ by adding the point probabilities for $k=0,1,\ldots,c$. The example below shows how to keep the terms and arithmetic organized.

Worked Example: At Most Two Successful Checks

Suppose a technician checks five independent components. Treat a check as a success if the component passes, and suppose each component independently passes with probability $0.30$. Let $X$ be the number of components that pass. Find the probability that at most two pass.

Check the model. There are a fixed five checks; each result is pass or fail; the checks are assumed independent; and the pass probability is $0.30$ for each component. Therefore $X\sim\operatorname{Bin}(5,0.30)$.

“At most two” means $X=0$, $X=1$, or $X=2$. The event includes exactly two; it is not restricted to counts below two. Using $1-p=0.70$,

\begin{aligned} P(X\le 2) &=P(X=0)+P(X=1)+P(X=2)\\ &=\binom{5}{0}(0.30)^0(0.70)^5 +\binom{5}{1}(0.30)^1(0.70)^4 +\binom{5}{2}(0.30)^2(0.70)^3\\ &=0.16807+0.36015+0.30870\\ &=0.83692. \end{aligned}

Thus the probability that at most two of the five components pass is $0.8369$, or about $83.69\%$. This is the model-based probability of that event under the stated assumptions, not a guarantee about the outcome of any particular set of checks.

Calculating an “At Least” Probability

“At least $c$” includes $c$ and every larger possible count, through $n$. For a binomial count, the largest possible value is the number of trials, so the sum must end at $n$. When the event contains only a few high counts, adding those terms directly is a straightforward way to calculate the probability.

Worked Example: At Least Six Successful Trials

In a hypothetical sequence of eight independent trials, suppose the probability of success on each trial is $0.25$. Let $X$ count the successes. Find the probability of at least six successes.

Check the model. The number of trials is fixed at eight, each trial has success or failure as its outcome, the trials are assumed independent, and the success probability is constant at $0.25$. Hence $X\sim\operatorname{Bin}(8,0.25)$.

“At least six” includes $X=6,7,8$, since eight is the maximum possible count. Add all three point probabilities:

\begin{aligned} P(X\ge 6) &=P(X=6)+P(X=7)+P(X=8)\\ &=\binom{8}{6}(0.25)^6(0.75)^2 +\binom{8}{7}(0.25)^7(0.75) +\binom{8}{8}(0.25)^8\\ &=0.003845215+0.000366211+0.000015259\\ &\approx 0.004226685. \end{aligned}

The probability is approximately $0.004227$, or $0.4227\%$. The event is rare under this model, but its probability is positive because six, seven, and eight successes are possible. Each term represents a different count and must be included once.

Calculating a Probability Between Two Counts

For a statement such as “between $a$ and $b$, inclusive,” include both endpoint counts and every integer count between them. If the wording does not say “inclusive,” pay careful attention to its exact meaning: “more than $a$” excludes $a$, while “at least $a$” includes it. Translating the words into inequalities before calculating helps prevent endpoint errors.

Worked Example: Between Three and Five Successes, Inclusive

Suppose a hypothetical process produces ten independent items, each of which meets a stated quality criterion with probability $0.40$. Let $X$ be the number that meet the criterion. Find the probability that between three and five items meet it, inclusive.

Check the model. There are ten fixed trials, each outcome is either meeting or not meeting the criterion, the outcomes are assumed independent, and the probability of meeting the criterion is $0.40$ for every item. Thus $X\sim\operatorname{Bin}(10,0.40)$.

“Between three and five, inclusive” means $3\le X\le5$, so the included counts are $3,4,5$. With $1-p=0.60$:

\begin{aligned} P(3\le X\le5) &=P(X=3)+P(X=4)+P(X=5)\\ &=\binom{10}{3}(0.40)^3(0.60)^7 +\binom{10}{4}(0.40)^4(0.60)^6 +\binom{10}{5}(0.40)^5(0.60)^5\\ &=0.214990848+0.250822656+0.200658125\\ &\approx0.666471629. \end{aligned}

Therefore, the probability that between three and five items meet the criterion, including both endpoints, is approximately $0.6665$, or $66.65\%$. The calculation includes exactly three and exactly five, as well as exactly four.

A Reliable Calculation Process

A short, consistent process reduces both modeling and arithmetic errors. In particular, write the event in inequality form before substituting into the binomial formula. This makes the included values visible and gives a clear way to check the summation limits.

Confirm the binomial model.
Identify the fixed number of trials, the two outcomes, independence, and the common success probability.

Translate the wording.
Write the event as an inequality, such as $X\le c$, $X\ge c$, or $a\le X\le b$.

List included integer counts.
Check the endpoints and the possible range $0,\ldots,n$.

Calculate and add.
Use the binomial point-probability formula for each included count, then add the terms and report the result in context.

Common Mistakes and Why the Details Matter

Cumulative probabilities answer questions about a collection of possible outcomes, not a single exact outcome. A value that is close to a cutoff can still be important: omitting one endpoint changes the event, and omitting an interior count leaves out one of its disjoint outcomes. These details matter whenever a probability is used to describe how likely a range of counts is under a model.

Dropping an endpoint. The incorrect reasoning is that “at most three” means counts below three. The correction is $X\le3$, which includes 0, 1, 2, and 3.
Adding too many values. The incorrect reasoning is to include counts below a lower bound in an event described as “at least.” The correction is to translate the phrase first: $X\ge c$ includes $c,c+1,\ldots,n$, and nothing below $c$.
Skipping an interior count. For $2\le X\le5$, the included counts are 2, 3, 4, and 5. Adding only the endpoint probabilities does not give the probability of the whole interval.
Using the wrong upper limit. An “at least” sum ends at $n$, the largest possible count, not at $n-1$.
Rounding each term too early. Keep several decimal places in the point probabilities, add them, and round the final probability. Early rounding can noticeably affect a sum with many terms.

A useful final check is to compare the terms you added with the event you wrote. For $P(X\le c)$, the list should start at zero and finish at $c$. For $P(X\ge c)$, it should start at $c$ and finish at $n$. For an inclusive interval, it should contain every integer from its lower endpoint through its upper endpoint.

Check Your Understanding

For each question, identify the included count values before deciding how to calculate the probability.

If $X\sim\operatorname{Bin}(7,0.20)$, which values of $X$ are included in the event “at most three successes”?
If $X\sim\operatorname{Bin}(9,0.50)$, write a sum of point probabilities for “at least seven successes.”
A binomial count is described as between two and five successes, inclusive. Write the event as an inequality and list the included integer counts.
Explain why the events $X=2$ and $X=3$ can have their probabilities added to find $P(X=2\text{ or }X=3)$.
A student calculates $P(X\le4)$ by adding $P(X=1),P(X=2),P(X=3),P(X=4)$. What count has been omitted, and how should the sum be corrected?