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When does a setting follow a binomial distribution, and how do we compute its probabilities?

Topic 4.10 Introduction to the Binomial Distribution: identify binomial settings (BINS conditions) and use the binomial probability formula to find the probability of a given number of successes in a fixed number of trials.

A focused answer to AP Statistics Topic 4.10, on the binomial setting (the BINS conditions), the binomial probability formula, and computing exact and cumulative binomial probabilities, with full worked calculations.

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  1. What this topic is asking
  2. The binomial setting
  3. The binomial probability formula
  4. Cumulative probabilities and "at least one"
  5. Recognizing binomial on the exam
  6. Try this

What this topic is asking

The College Board (Topic 4.10) wants you to recognize a binomial setting by its conditions and use the binomial probability formula to find the probability of a given number of successes in a fixed number of trials.

The binomial setting

The first job in any binomial question is to check these conditions, because the formula is only valid when they hold. Flipping a coin nn times, guessing nn multiple-choice questions, or inspecting nn items for defects (with replacement or from a large population) all fit. The fixed number of trials is what distinguishes binomial from the geometric setting (Topic 4.12), where you instead wait for the first success.

The binomial probability formula

The formula has three pieces, each with a meaning. px(1−p)n−xp^x (1-p)^{n-x} is the probability of one specific sequence with xx successes (by independence, you multiply). The binomial coefficient (nx)\binom{n}{x} counts how many such sequences there are, because the successes can fall in different positions. Multiplying them gives the total probability of getting exactly xx successes in any order. Forgetting the coefficient, a very common slip, undercounts the arrangements and gives the wrong answer.

Cumulative probabilities and "at least one"

Many questions ask not for an exact count but for a range: "at least 33," "at most 22," "fewer than 55." These are sums of binomial terms over the relevant values of xx. "At most 22" is P(0)+P(1)+P(2)P(0) + P(1) + P(2); "more than 11" is everything from 22 up. The most efficient special case is "at least one," which is the complement of "none": P(X≥1)=1−P(X=0)=1−(1−p)nP(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n, a single term rather than a long sum. Whenever a range has many values, check whether its complement has fewer, "at least 11" has a one-term complement, and "at most 11" is shorter than "at least 22" for large nn. A graphing calculator computes individual (binompdf) and cumulative (binomcdf) binomial probabilities directly, but the exam still expects you to set up the correct formula and identify nn, pp, and xx, and to recognize when the complement is the smart route. Translating the wording into the right set of values and choosing the efficient computation is the real skill here.

Recognizing binomial on the exam

Because the formula only applies when the conditions hold, the exam often tests whether you can tell binomial from non-binomial. A setting fails the binomial conditions if the number of trials is not fixed (you keep going until something happens, that is geometric), if the success probability changes from trial to trial (such as drawing without replacement from a small population, which breaks the "same pp" and independence conditions), or if outcomes are not naturally two-valued. A useful guideline for sampling without replacement: it is approximately binomial when the sample is no more than about 10%10\% of the population, because then pp barely changes between draws (the "10%10\% condition"). Being able to state the BINS conditions, verify them in context, and spot when one fails is exactly what separates a full-credit binomial answer from a misapplied formula, and it sets up Topic 4.11, which adds the binomial's mean and standard deviation.

Try this

Q1. State the four binomial (BINS) conditions. [2 points]

  • Cue. Binary outcomes; Independent trials; Number of trials fixed in advance; Same success probability on every trial.

Q2. For X∼B(10,0.2)X \sim B(10, 0.2), write the expression for P(X≥1)P(X \ge 1) using the complement. [1 point]

  • Cue. P(X≥1)=1−P(X=0)=1−(0.8)10P(X \ge 1) = 1 - P(X = 0) = 1 - (0.8)^{10}.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2019 (style)1 marksSection I (multiple choice). A fair coin is flipped 55 times. What is the probability of exactly 22 heads? (A) (52)(0.5)2(0.5)3\binom{5}{2}(0.5)^2(0.5)^3 (B) (0.5)2(0.5)^2 (C) (52)(0.5)2\binom{5}{2}(0.5)^2 (D) 2/52/5
Show worked answer →

The correct answer is (A).

This is binomial with n=5n = 5, p=0.5p = 0.5, x=2x = 2. The binomial formula is P(X=x)=(nx)px(1−p)n−x=(52)(0.5)2(0.5)3P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} = \binom{5}{2}(0.5)^2(0.5)^3.

(B) ignores the other flips and the number of arrangements. (C) omits the failure factor (0.5)3(0.5)^3. (D) is not a binomial probability. The full formula with the binomial coefficient is (A).

AP 2022 (style)4 marksSection II (free response). A multiple-choice quiz has 88 questions, each with 44 options; a student guesses every answer. Let XX be the number correct. (a) Explain why XX is binomial, checking the conditions. (b) Find the probability of exactly 33 correct. (c) Find the probability of at least 11 correct, and interpret in context.
Show worked answer →

A 4-point binomial question.

(a) (1 point) Binomial conditions hold: a fixed number of trials (n=8n = 8), each question is a success/failure (correct or not), constant success probability p=0.25p = 0.25, and questions are independent.
(b) (1 point) P(X=3)=(83)(0.25)3(0.75)5=56(0.015625)(0.2373)≈0.2076P(X = 3) = \binom{8}{3}(0.25)^3(0.75)^5 = 56(0.015625)(0.2373) \approx 0.2076.
(c) (2 points) Use the complement: P(X≥1)=1−P(X=0)=1−(0.75)8=1−0.1001=0.8999P(X \ge 1) = 1 - P(X = 0) = 1 - (0.75)^8 = 1 - 0.1001 = 0.8999 (1 point); interpret: there is about a 90%90\% chance the student gets at least one question right by guessing (1 point, in context).

Markers reward checking the binomial conditions, the exact binomial probability for X=3X = 3, and the complement for "at least one" with a contextual interpretation.

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