The Binomial Distribution — AP® Statistics Lesson

Part 1 · Hook

Why this matters

A coffee shop says 30% of its customers order with its app. You watch 10 random customers and 7 of them use it. Is the shop wrong, or did you see a fluke? When a chance process repeats the same yes-or-no trial a fixed number of times, one formula gives every probability you need to decide.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\binom{5}{2}$ ?

10
20
7
5

Show the answer

$\binom{5}{2}$ $= \frac{5!}{2!\,3!}$ $= 10$ .

Correct: 10:
20:
7:
5:

2. Events are independent when

knowing one happened does not change the probability of the other
they cannot happen together
their probabilities add to 1
they have the same probability

Show the answer

That is the definition: $P(A \mid B) = P(A)$ .

Correct: knowing one happened does not change the probability of the other:
they cannot happen together:
their probabilities add to 1:
they have the same probability:

Part 4 · See it

See it first

Probability histogram of the binomial distribution with n = 10 and p = 0.3. The bars for 4 through 10 are shaded; together they are P(X ≥ 4) = 1 − P(X ≤ 3) = 0.350. — The binomial distribution with n = 10 and p = 0.3. The shaded bars, 4 through 10, add to P(X ≥ 4) = 0.350. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A trial has two outcomes and repeats a fixed number of times, independently, with the same pthe count of successes X is binomial with parameters n and p
One order of k successes and n − k failures has probability $p^k(1 - p)^{n - k}$ and there are $\binom{n}{k}$ such orders, so $P(X = k)$ $= \binom{n}{k}p^k(1 - p)^{n - k}$
An event like "at least 4" covers several valuesadd them, or use the complement: $P(X \ge 4) = 1 - P(X \le 3)$
The probability of a result as extreme as the one seen is very small under a claimthe result gives reason to doubt the claim

Part 6 · Key ideas

Key ideas

A binomial setting: two outcomes per trial, independent trials, a fixed number n of trials, the same success probability p on every trial.
The binomial distribution: $P(X = k)$ $= \binom{n}{k}p^k(1 - p)^{n - k}$ for $k = 0, 1, \dots, n$ .
Show work: name the distribution and parameters, the boundary and the direction: "X is binomial, $n = 10$ , $p = 0.3$ ; $P(X \ge 4)$ $= 1 - P(X \le 3)$ $= 0.350$ ."
Mean $\mu_X = np$ and standard deviation $\sigma_X = \sqrt{np(1 - p)}$ .
Translate carefully: "fewer than 3" is $P(X \le 2)$ , "at least 4" is $P(X \ge 4)$ .

Part 7 · Misconception

A common mistake

The wrong idea: $P(X \ge 4)$ is $1 - P(X \le 4)$ .

What actually happens: The complement of "4 or more" is "3 or fewer", so $P(X \ge 4) = 1 - P(X \le 3)$ . Subtracting $P(X \le 4)$ throws away the value 4 itself.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Guessing on a quiz

A quiz has 12 multiple-choice questions, each with 4 options and one right answer. A student who has not studied guesses at random on every question, independently. Let X = the number of questions the student gets right.

1. What is the probability that the student gets exactly 3 questions right?

0.25
0.649
0.258
0.001

Show the answer

X is binomial with $n = 12$ and $p = 0.25$ . $P(X = 3)$ $= \binom{12}{3}(0.25)^3(0.75)^9$ $= 220(0.015625)$ $(0.07508)$ $\approx 0.258$ .

0.25: That is the chance of guessing one question right, not of exactly 3 right out of 12.
0.649: That is $P(X \le 3)$ , the probability of 3 or fewer.
Correct: 0.258: Right: the binomial formula with n = 12, p = 0.25, k = 3.
0.001: That is one arrangement of 3 right and 9 wrong. Multiply by $\binom{12}{3} = 220$ arrangements.

2. What is the probability that the student gets at least 6 questions right?

0.1576
0.0401
0.9456
0.0544

Show the answer

$P(X \ge 6)$ $= 1 - P(X \le 5)$ $= 1 - 0.9456$ $= 0.0544$ , with X binomial, $n = 12$ , $p = 0.25$ .

0.1576: That is $P(X \ge 5)$ : subtracting $P(X \le 4)$ leaves 5 in.
0.0401: That is $P(X = 6)$ alone. At least 6 includes 7 through 12.
0.9456: That is $P(X \le 5)$ , the complement.
Correct: 0.0544: Right: 1 minus the cumulative probability up to 5.

3. What are the mean and standard deviation of X?

Mean 3, standard deviation 1.5
Mean 3, standard deviation 2.25
Mean 6, standard deviation 1.5
Mean 3, standard deviation 0.75

Show the answer

$\mu_X = np = 12(0.25) = 3$ questions and $\sigma_X$ $= \sqrt{np(1 - p)}$ $= \sqrt{12(0.25)(0.75)}$ $= \sqrt{2.25}$ $= 1.5$ questions.

Correct: Mean 3, standard deviation 1.5: Right: np and the square root of np(1 − p).
Mean 3, standard deviation 2.25: 2.25 is the variance; take its square root.
Mean 6, standard deviation 1.5: The mean is np = 3, not half of the 12 questions.
Mean 3, standard deviation 0.75: 0.75 is 1 − p, not the standard deviation.

Table

Survey phone calls

A polling firm dials 15 phone numbers chosen at random from a list of over a million numbers in a city. From long experience, 20% of calls are answered. Let Y = the number of the 15 calls that are answered (invented data).

The polling setup
Number of calls	Chance a call is answered	Y counts
15	0.20	answered calls

4. Why is it reasonable to model Y with a binomial distribution?

The 15 numbers are a random sample from the list, and 15 calls is a large enough number of trials
Each call is answered or not, n = 15 is fixed, p = 0.20 each time, calls are independent
About 20% of the 15 calls, which is 3 calls, will be answered
The firm keeps dialing new numbers until someone answers a call

Show the answer

Binomial setting: each call is answered or not (two outcomes), the number of calls is fixed at 15, the chance of an answer is 0.20 on each call, and calls to numbers chosen at random from a huge list are independent.

The 15 numbers are a random sample from the list, and 15 calls is a large enough number of trials: Random choice helps independence, but the fixed number, two outcomes and constant p are also needed.
Correct: Each call is answered or not, n = 15 is fixed, p = 0.20 each time, calls are independent: Right: all four conditions, in context.
About 20% of the 15 calls, which is 3 calls, will be answered: That is the mean, not a reason the distribution is binomial.
The firm keeps dialing new numbers until someone answers a call: The firm dials exactly 15 numbers; a fixed number of trials is one of the conditions.

5. What is the probability that exactly 2 of the 15 calls are answered?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Y is binomial with $n = 15$ , $p = 0.20$ . $P(Y = 2)$ $= \binom{15}{2}(0.20)^2(0.80)^{13}$ $= 105(0.04)$ $(0.80)^{13}$ $\approx 0.2309$ .

Answer: 0.2309

6. What is the probability that fewer than 2 of the calls are answered?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

"Fewer than 2" means 0 or 1: $P(Y \le 1)$ $= P(Y = 0)$ $+ P(Y = 1)$ $= (0.80)^{15}$ $+ 15(0.20)$ $(0.80)^{14}$ $\approx 0.1671$ .

Answer: 0.1671

Part 9 · Summary

Summary

When a yes-or-no trial repeats n times independently with the same success probability p, the number of successes is binomial:

P(X = k)

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

, with mean np and standard deviation

\sqrt{np(1 - p)}

. Name the distribution and its parameters, translate the event into an inequality, and use the complement for "at least" questions. A tiny probability under a claim casts doubt on the claim.

Part 10 · Up next

What comes next

Part 11 · Connections