Unit 2 · Topic 2.10 · about 30 minutes

The Binomial Distribution

Justify whether a count is a binomial random variable, then calculate and interpret its probabilities, mean and standard deviation in context.

Predict first

A quiz has 10 multiple-choice questions, each with 4 choices. A student guesses at random on every question. How likely is it that the student gets at least 6 right?

The binomial setting

Many random variables count successes, such as the free throws a player makes or the seeds that sprout. A binomial random variable X counts the number of successes in repeated trials when all four of these hold:

Two outcomes. Each trial ends in a success or a failure.
Independent trials. The result of one trial does not change the chances on any other.
A fixed number of trials, called $n$ , set before you start.
The same probability of success $p$ on every trial, so the probability of failure is $1 - p$ .

Some teachers call this list BINS, for binary, independent, number and same. Whatever you call it, check it in context before you use any binomial formula. "Success" just means the outcome being counted; it can be a defective phone or a missed shot.

Sort it

Is X a binomial random variable? Tap a card, then tap its bin. Each card explains itself once everything is placed.

Binomial

Not binomial

The binomial probability function

If X is binomial with $n$ trials and probability of success $p$ , then

P (X = x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x}, x = 0, 1, 2, \dots, n

This is the binomial probability function. The factor $p^{x} (1 - p)^{n - x}$ is the probability of one particular arrangement of $x$ successes and $n - x$ failures, such as success, success, failure, success. The factor $(\binom{n}{x})$ , read "n choose x", counts how many such arrangements there are. Your calculator finds it with nCr.

Two calculator commands do most of the work. binompdf(n, p, x) gives $P (X = x)$ . binomcdf(n, p, x) gives the cumulative probability $P (X \leq x)$ .

Worked exampleFree throws in tonight's game

A basketball player makes 75% of her free throws. Tonight she will shoot 8 free throws. Assume her shots are independent, and let X be the number she makes. Find $P (X = 6)$ and $P (X \geq 6)$ .

Check the binomial setting. Each shot is a make or a miss. There are a fixed 8 shots. The shots are independent, as stated. The chance of a make is 0.75 on every shot. So X is binomial with $n = 8$ and $p = 0.75$ .
Exactly 6, by the formula. $P (X = 6) = (\binom{8}{6}) (0.75)^{6} (0.25)^{2} = 28 (0.17798) (0.0625) \approx 0.3115$ . On the calculator: binompdf(8, 0.75, 6).
At least 6, by the complement. "At least 6" means 6, 7 or 8, which is the complement of "at most 5": $P (X \geq 6) = 1 - P (X \leq 5) \approx 1 - 0.3215 = 0.6785$ . On the calculator: 1 minus binomcdf(8, 0.75, 5).
Interpret in context. There is about a 31% chance that she makes exactly 6 of her 8 free throws, and about a 68% chance that she makes 6 or more.

Answer.

$P (X = 6) \approx 0.3115$ and $P (X \geq 6) \approx 0.6785$ .

Binomial distribution with n = 8 and p = 0.75

The bars for 6, 7 and 8 together make up $P (X \geq 6) \approx 0.68$ . With $p$ this close to 1, the distribution is skewed to the left.

Mean and standard deviation

A binomial random variable has

μ_{X} = n p σ_{X} = \sqrt{n p (1 - p)}

For the free throws, $μ_{X} = 8 (0.75) = 6$ and $σ_{X} = \sqrt{8 (0.75) (0.25)} = \sqrt{1.5} \approx 1.22$ . In context: if she shot many sets of 8 free throws, she would make 6 per set on average, and the number she makes in a set would typically differ from 6 by about 1.22. These are the same mean and standard deviation as in 2.9 Parameters of Random Variables. The binomial formulas are shortcuts that work for this one family of distributions.

Estimating with a simulation

A binomial probability can also be estimated by simulation, as in 2.3 Estimating Probabilities Using Simulation. To model one free throw, generate a random integer from 1 to 4 and let 1, 2 or 3 stand for a make, which gives probability 0.75. One trial is 8 such integers, and you record whether 6 or more of them are makes. In 1,000 simulated sets of 8 shots, 671 sets had at least 6 makes, so the estimate is $\frac{671}{1000} = 0.671$ , close to the exact 0.6785.

Lab

Binomial Explorer

Set n = 8 and p = 0.75, then choose P(X ≥ k) under Find with k = 6 to shade "at least 6" and check the example. Then slide p. The distribution is skewed left when p is near 1, skewed right when p is near 0, and symmetric when p = 0.5.

Open the full Binomial Explorer lab

Check your understanding

A teacher picks 4 students at random, without replacement, from a class of 12 in which 5 students have a pet. Let X be the number of chosen students who have a pet. Is X a binomial random variable? Choose the response with the correct answer and a valid reason.

Seeds from a packet germinate with probability 0.85, independently of each other. You plant 12 seeds. What is the probability that exactly 10 of them germinate? Round to four decimal places.

Your answer

A survey firm reaches a live person on 20% of its calls, independently from call to call. Let X be the number of people reached in 15 calls. Which expression gives the probability of reaching at least 3 people?

A store offers its credit card at checkout, and 12% of customers accept, independently of each other. For the next 50 customers, the number who accept, X, has mean $μ_{X} = 6$ and standard deviation $σ_{X} \approx 2.30$ . Which is the best interpretation of $σ_{X}$ ?

A delivery company claims that 90% of its packages arrive on time. You track 20 randomly chosen packages, and only 14 arrive on time. Assume packages arrive on time independently. If the claim is true, $P (X \leq 14) \approx 0.011$ . What does this suggest?

Practice

Practice until it is automatic

New numbers every time. Each one is checked the moment you answer, with the full working shown.

Binomial probabilities practice page

Course alignment, for teachers

AP Statistics topic 2.10, Unit 2: Probability, Random Variables, and Probability Distributions.

Skill 3.C: Calculate and estimate expected counts, percentages, probabilities, and intervals.
Skill 3.D: Calculate means, standard deviations, and parameters for probability distributions.
Skill 4.B: Justify a claim based on statistical calculations and results.
Skill 4.D: Interpret statistical calculations and results to assess meaning or a claim.