binompdf & binomcdf on TI-84 — Binomial Probability Guide

Right after normalcdf and invNorm, the binomial commands are the most-used distribution functions on the AP Statistics exam. This guide covers exactly when to use binompdf vs binomcdf, how to handle "at least" and "more than" questions, and how to write your answers so AP graders give full credit.

When is a situation binomial?

A binomial setting has four conditions (the classic BINS check):

1. Binary — each trial is a success or failure
2. Independent — trials don't affect each other
3. Number — a fixed number of trials, n
4. Same probability — the success probability p is constant

If all four hold, X ~ B(n, p) and the TI-84 can compute any probability about X.

binompdf vs binomcdf — the critical difference

pdf = exactly one value. cdf = a running total from 0 up to x. Mixing these up is the single most common binomial mistake on exams.

How to access the commands

1. Press 2nd VARS to open the DISTR menu
2. Scroll down to A: binompdf( or B: binomcdf(
3. On the TI-84 Plus CE, a Stat Wizard appears with three prompts: trials (n), p, and x value
4. Fill them in, highlight Paste, and press ENTER twice

Tip: scrolling up from the top of the DISTR menu wraps around to the bottom, which reaches A: and B: faster.

Worked Example 1: Coin flips (exactly)

Question: You flip a fair coin 10 times. What is the probability of getting exactly 4 heads?

Here X ~ B(10, 0.5) and we want P(X = 4) — the word "exactly" means binompdf.

Keystrokes:

2nd → VARS → A
binompdf(10, 0.5, 4)

Result: 0.205078125

AP-style answer: "Let X = the number of heads in 10 flips. X ~ B(10, 0.5). binompdf(10, 0.5, 4) = 0.2051, so there is about a 20.5% probability of exactly 4 heads."

Worked Example 2: Coin flips (at most)

Question: Same 10 flips — what is the probability of getting at most 3 heads?

"At most 3" means P(X ≤ 3) — that's binomcdf.

Keystrokes:

binomcdf(10, 0.5, 3)

Result: 0.171875

This equals P(X=0) + P(X=1) + P(X=2) + P(X=3) added together — the cdf does the summing for you.

Worked Example 3: Free throws (exactly, at least, more than)

Question: A basketball player makes 60% of her free throws. She attempts 10. Let X ~ B(10, 0.6).

(a) P(exactly 7 makes):

binompdf(10, 0.6, 7)

Result: 0.2149908480 → about 0.2150

(b) P(at least 8 makes): "At least 8" is the complement of "7 or fewer":

1 - binomcdf(10, 0.6, 7)

Result: 0.1672897536 → about 0.1673

(c) P(more than 8 makes): "More than 8" means 9 or 10, the complement of "8 or fewer":

1 - binomcdf(10, 0.6, 8)

Result: 0.0463574016 → about 0.0464

Translation table for word problems

Memorize the last two rows: "at least k" subtracts the cdf at k−1, while "more than k" subtracts the cdf at k. The difference is whether k itself is included.

AP Exam Tip: show your setup

Calculator answers with no work earn no credit. For every binomial question, write three things:

1. Define the variable and distribution: "Let X = the number of made free throws. X ~ B(10, 0.6)."
2. State the calculator command with its inputs: "1 − binomcdf(10, 0.6, 7)"
3. Report the result in context: "= 0.1673, so there is about a 16.7% chance she makes at least 8 of 10 free throws."

Graders explicitly look for n, p, and the boundary value — writing the full command shows all three.

Common errors

• Entering p as a percent. binompdf(10, 60, 7) uses p = 60, which is not a valid probability and triggers ERR:DOMAIN. Always convert: 60% → 0.6. See the ERR:DOMAIN fix guide if you hit this message.
• Using binomcdf for "exactly." binomcdf(10, 0.6, 7) = 0.8327, which is P(X ≤ 7), not P(X = 7) = 0.2150 — a wildly different answer.
• Off-by-one on complements. For "at least 8," subtract the cdf at 7, not at 8.

Practice right now

You don't need the physical handheld to drill these — open the free online TI-84 calculator (a full TI-84 Plus CE in your browser, no download) and try it yourself: compute binompdf(10, 0.5, 4) and 1 - binomcdf(10, 0.6, 7) and check that you get 0.2051 and 0.1673. Five minutes of practice now saves easy points on exam day.