binompdf vs binomcdf: which one do I use?

Q: Which command do I use for 'at least'?

Neither one on its own. Use the complement: P(X ≥ k) = 1 - P(X ≤ k-1), which is 1 - binomcdf(n, p, k-1). For at least 4 out of 10 with p = 0.3, that is 1 - binomcdf(10, 0.3, 3) = 1 - 0.6496 = 0.3504.

Q: Does binomcdf include the x value I type?

Yes. binomcdf(n, p, x) is P(X ≤ x), so the count x itself is part of the total. For fewer than x, which excludes it, drop the boundary by one and use binomcdf(n, p, x-1).

Q: What if my calculator does not have binompdf and binomcdf?

Use the formula P(X = x) = C(n, x) p^x (1-p)^n-x, which is on the provided AP formula sheet, and add terms for a cumulative probability. Any calculator with a factorial or combination key can do it, and the arithmetic is short for small n.

By Jude Wallis · Published July 30, 2026

binompdf(n, p, x) returns the probability of exactly x successes. binomcdf(n, p, x) returns the cumulative probability of x or fewer. Match the wording: exactly means pdf, at most means cdf, and at least k means 1 - binomcdf(n, p, k-1).

AP Statistics: Unit 2 (topics 2.10 The Binomial Distribution). Binomial probabilities are Unit 2 Topic 2.10 of the Fall 2026 AP Statistics course, where calculating probabilities for a binomial distribution is a stated learning objective and the binomial probability function appears on the provided formula sheet.

pdf means exactly, cdf means at most

The difference comes down to one word. binompdf(n, p, x) returns the probability of exactly $x$ successes, written $P(X = x)$ . binomcdf(n, p, x) returns the probability of x or fewer successes, written $P(X \le x)$ .

The c in binomcdf stands for cumulative, and cumulative means the command has already added every outcome from 0 through $x$ . So binomcdf(10, 0.3, 3) is the single-step version of $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ .

In both commands $X$ is the random variable counting successes, $n$ is the fixed number of trials, and $p$ is the probability of success on a single trial. If those pieces do not describe your situation, neither command applies, so check the setup first with the binomial probability guide.

Where the commands live and what order to type

Both sit in the distribution menu, reached with 2nd VARS. The menu is labeled DISTR, and the binomial entries sit below the normal and t entries, so you will scroll. The normal commands in the same menu are covered in normalcdf and invNorm on the TI-84.

Item numbers differ across TI-84 models, so scroll to the name rather than memorizing a number. binompdf and binomcdf sit next to each other in the list, which is exactly why the wrong one gets selected.

The argument order is trials, then the success probability, then the count: binompdf(n, p, x). Some calculators open a fill-in screen with lines for the number of trials, $p$ , and the x value, and others expect the three values inside parentheses separated by commas. The order does not change.

Leaving off the final count returns a whole list of probabilities, one for each $x$ from 0 to $n$ , instead of a single number. That is handy for building a probability distribution table and confusing if you did not mean to do it.

The decision table

Translate the wording into symbols first, then read the command off the table. Getting the boundary right matters more than the typing.

Wording	Symbols	Command
Exactly $k$	$P(X = k)$	binompdf(n, p, k)
At most $k$ , $k$ or fewer, no more than $k$	$P(X \le k)$	binomcdf(n, p, k)
Fewer than $k$ , less than $k$	$P(X \le k-1)$	binomcdf(n, p, k-1)
At least $k$ , $k$ or more	$P(X \ge k)$	1 - binomcdf(n, p, k-1)
More than $k$ , greater than $k$	$P(X > k)$	1 - binomcdf(n, p, k)
Between $a$ and $b$ , inclusive	$P(a \le X \le b)$	binomcdf(n, p, b) - binomcdf(n, p, a-1)

Two pairs of rows cause most of the trouble. At most includes $k$ itself while fewer than excludes it, so the two differ by one whole term. The same one-term gap separates at least from more than.

Why at least needs the complement

There is no calculator command for at least, so you build it from the cdf. At least $k$ is the exact opposite of at most $k-1$ , because a count either sits below $k$ or sits at $k$ and above, with nothing in between:

P(X \ge k) = 1 - P(X \le k-1)

On a calculator that is 1 - binomcdf(n, p, k-1). The $k-1$ is where the errors live. Subtracting binomcdf(n, p, k) instead would strip out the $P(X = k)$ term the question told you to keep.

Test the rule on the smallest case whenever you doubt it. At least 1 should be $1 - P(X = 0)$ , and the rule gives $1 - P(X \le 0)$ , which is the same thing because $P(X \le 0) = P(X = 0)$ . If your version of the rule fails that test, the offset is wrong. The at least one guide works through that case in detail.

Ranges between two counts

For between $a$ and $b$ inclusive, subtract two cumulative values: binomcdf(n, p, b) - binomcdf(n, p, a-1). Subtracting binomcdf(n, p, a) instead would delete $P(X = a)$ , which the word inclusive told you to keep.

Read the endpoints carefully, because ordinary English is vague here. Look for inclusive, or for from 3 to 6, or for more than 3 but fewer than 6, to decide whether the endpoints belong in the range.

On a short range you can skip the subtraction and add exact probabilities instead. For $n = 10$ and $p = 0.3$ , $P(3 \le X \le 4)$ is binompdf(10, 0.3, 3) + binompdf(10, 0.3, 4) = $0.2668 + 0.2001 = 0.4669$ .

Mistakes that turn a correct setup into a wrong answer

Using binompdf for a range. binompdf answers only the exactly question. If the wording is at most, at least, fewer than, or more than, binomcdf has to appear somewhere in your expression.
Using binomcdf for exactly. binomcdf(10, 0.3, 3) is 0.6496, while binompdf(10, 0.3, 3) is 0.2668. A cumulative value is always at least as large as the matching exact value, so a surprisingly big probability is a hint you grabbed the cdf.
Going off by one at the boundary. At least 4 uses binomcdf(10, 0.3, 3), not binomcdf(10, 0.3, 4). Writing the symbol statement before you type prevents this one.
**Swapping $n$ and $p$ .** Trials come first, then the probability. Because $p$ sits between 0 and 1 and $n$ is a whole number, a calculator error or an absurd result usually traces back here.
Using a binomial when the setup is not binomial. Drawing without replacement from a small population breaks independence, and counting trials until the first success is geometric rather than binomial.

Checking a calculator answer against the formula

The binomial probability function is on the provided AP formula sheet, so any binompdf value can be confirmed by hand:

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

The coefficient $\binom{n}{x}$ , read n choose x, counts the arrangements that produce $x$ successes. The factor $p^x$ covers the successes and $(1-p)^{n-x}$ covers the remaining failures.

A cumulative value takes more arithmetic but no new ideas, since binomcdf(n, p, x) is that same formula evaluated at $0, 1, \ldots, x$ and summed. Both worked examples below do it term by term, and the binomial probability calculator takes the same three inputs if you want a second opinion.

One more check costs nothing. The mean of a binomial is $\mu_X = np$ (mu sub X, the expected count), so for $n = 10$ and $p = 0.3$ the average number of successes is 3. Probabilities for counts near 3 should be sizeable, and probabilities for counts near 10 should be tiny.

How this fits the Fall 2026 AP course

The binomial distribution is Topic 2.10 in the Fall 2026 AP Statistics course, with the first exam in May 2027. It sits in Unit 2, Probability, Random Variables, and Probability Distributions, which carries 15 to 25 percent of the multiple-choice section. Calculating probabilities for a binomial distribution is a stated learning objective, and the probability function appears on the provided formula sheet.

The Course and Exam Description also states that a graphing calculator with statistical capabilities is expected for both sections of the exam, so these commands are available throughout. What they do not do is write your reasoning for you.

On free response, the command alone leaves out the setup. Define the random variable, say that it is binomial with the values of $n$ and $p$ , write the probability statement in symbols, then give the number and a sentence in context. A grader reading only binomcdf(10, 0.3, 3) cannot tell whether you understood the boundary or guessed it.

For the structure of a written response, see the AP Statistics FRQ guide and reading calculator output. The official course page carries the framework itself.

Exactly 3 successes with n = 10 and p = 0.3

An inspector checks 10 independently produced items, each defective with probability 0.3. Let $X$ count the defective items. Find $P(X = 3)$ .

Check the setup: two outcomes per item, a fixed 10 trials, the same $p = 0.3$ each time, and independent items, so $X$ is binomial with $n = 10$ and $p = 0.3$ .
The word exactly points to binompdf, so the command is binompdf(10, 0.3, 3).
The calculator returns 0.2668.
Verify with the formula. The coefficient is $\binom{10}{3} = \frac{10!}{3!\,7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$ .
Success factor: $p^x = 0.3^3 = 0.027$ .
Failure factor: $(1-p)^{n-x} = 0.7^7 = 0.0823543$ . Building it up, $0.7^2 = 0.49$ , $0.7^3 = 0.343$ , $0.7^4 = 0.2401$ , $0.7^5 = 0.16807$ , $0.7^6 = 0.117649$ , $0.7^7 = 0.0823543$ .
Multiply the three pieces: $120 \times 0.027 = 3.24$ , then $3.24 \times 0.0823543 = 0.266827932$ .
Round to four decimals: 0.2668, matching binompdf.

$P(X = 3) = 0.2668$ , about a 26.68% chance of exactly 3 defective items. The command is binompdf(10, 0.3, 3).

At least 4 successes with n = 10 and p = 0.3

Same setting: 10 independent items, each defective with probability 0.3, and $X$ counts the defectives. Find $P(X \ge 4)$ .

Translate the wording. At least 4 means $X \ge 4$ , so the complement is $X \le 3$ , and $P(X \ge 4) = 1 - P(X \le 3)$ .
Apply the rule with $k = 4$ , so $k - 1 = 3$ : the command is 1 - binomcdf(10, 0.3, 3).
The calculator gives binomcdf(10, 0.3, 3) = 0.6496, so $P(X \ge 4) = 1 - 0.6496 = 0.3504$ .
Verify the cumulative value term by term. $P(X = 0) = \binom{10}{0}(0.3)^0(0.7)^{10} = 1 \times 1 \times 0.0282475249 = 0.0282475249$ .
$P(X = 1) = \binom{10}{1}(0.3)^1(0.7)^9 = 10 \times 0.3 \times 0.040353607 = 3 \times 0.040353607 = 0.121060821$ .
$P(X = 2) = \binom{10}{2}(0.3)^2(0.7)^8 = 45 \times 0.09 \times 0.05764801 = 4.05 \times 0.05764801 = 0.2334744405$ .
$P(X = 3) = 0.266827932$ , computed in the previous example.
Add the four terms: $0.0282475249 + 0.121060821 = 0.1493083459$ , then $0.1493083459 + 0.2334744405 = 0.3827827864$ , then $0.3827827864 + 0.266827932 = 0.6496107184$ . That matches binomcdf(10, 0.3, 3) = 0.6496.
Subtract from 1: $1 - 0.6496107184 = 0.3503892816$ , which rounds to 0.3504.
Sanity check: the mean is $np = 10 \times 0.3 = 3$ , so at least 4 asks for the chance of landing above average. A value near 0.35 is reasonable, while a value above 0.5 would not be.

$P(X \ge 4) = 1 - 0.6496 = 0.3504$ , about a 35.04% chance of at least 4 defective items. The command is 1 - binomcdf(10, 0.3, 3).

Frequently asked questions

Which command do I use for 'at least'?

Neither one on its own. Use the complement: $P(X \ge k) = 1 - P(X \le k-1)$ , which is 1 - binomcdf(n, p, k-1). For at least 4 out of 10 with $p = 0.3$ , that is 1 - binomcdf(10, 0.3, 3) = 1 - 0.6496 = 0.3504.

Does binomcdf include the x value I type?

Yes. binomcdf(n, p, x) is $P(X \le x)$ , so the count $x$ itself is part of the total. For fewer than $x$ , which excludes it, drop the boundary by one and use binomcdf(n, p, x-1).

Is binompdf really giving me a probability?

Yes. The name borrows pdf from continuous distributions, where such a function gives a density rather than a probability. For a discrete count like a binomial, binompdf(n, p, x) returns an actual probability, $P(X = x)$ .

What if my calculator does not have binompdf and binomcdf?

Use the formula $P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$ , which is on the provided AP formula sheet, and add terms for a cumulative probability. Any calculator with a factorial or combination key can do it, and the arithmetic is short for small $n$ .

Is the binomial distribution on the Fall 2026 AP exam?

Yes. It is Unit 2 Topic 2.10, and Unit 2 carries 15 to 25 percent of the multiple-choice section. The binomial probability function, the mean $np$ , and the standard deviation $\sqrt{np(1-p)}$ all appear on the provided formula sheet.