AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/2/2-10)
Unit 2 · Topic 2.10
2.10 The Binomial Distribution
A binomial random variable counts successes in a fixed number of independent trials that each have the same chance of success. You'll learn to check whether a setting is binomial, find probabilities with the formula or technology, and compute and interpret the mean np and standard deviation √(np(1 − p)).
Key terms
- binomial random variable
- success and failure
- independent trials
- binomial probability
- mean np and standard deviation √(np(1 − p))
Is it binomial?
X is binomial if all four of these hold:
- Binary: each trial has two outcomes, success or failure.
- Independent: one trial's result doesn't affect another's.
- Number: there's a fixed number of trials, n.
- Same probability: the chance of success, p, is the same on every trial.
Sampling without replacement
Drawing without replacement makes trials dependent, because each draw changes what's left. If the sample is small compared with the population (no more than 10% of it), the change is tiny and you can treat the trials as independent. This is the 10% condition, and you'll use it again throughout inference.
Drawing 13 cards from a 52-card deck and counting hearts is not binomial: 13 is 25% of 52, and the chance of a heart changes noticeably after each draw. Surveying 50 people out of a city of 80,000 is close enough to binomial.
Binomial probabilities
P(X = x) = C(n, x) · pˣ · (1 − p)ⁿ⁻ˣ for x = 0, 1, 2, …, n. Here C(n, x) = n! ÷ [x!(n − x)!] counts the ways to arrange x successes among n trials. The formula is on the formula sheet.
On a calculator or in Desmos, use binompdf(n, p, x) for exactly x and binomcdf(n, p, x) for P(X ≤ x). For "at least x," use 1 − P(X ≤ x − 1).
You can also estimate binomial probabilities with a simulation, as in 2.3.
Mean and standard deviation
For a binomial variable, μX = np and σX = √(np(1 − p)). If 30% of customers use a coupon and you watch 10, you expect μX = 10(0.3) = 3 coupon users, with standard deviation √(10 × 0.3 × 0.7) = √2.1 ≈ 1.45.
Interpret in context: "Over many groups of 10 customers, the average number using a coupon would be 3, and the count typically varies from 3 by about 1.45."
The shape of a binomial distribution depends on n and p. When p = 0.5 it's symmetric. When p is small it's skewed right, and when p is large it's skewed left. As n grows, the shape gets closer to normal, which is why normal models work for proportions in Unit 3.
You won't be tested on the geometric distribution (counting trials until the first success). It was removed from the course.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Exactly, at most and at least
30% of a store's customers use a coupon, independently. For 10 randomly chosen customers, find P(X = 3), P(X ≤ 2) and P(X ≥ 3).
Show the solutionHide the solution
- Step 1: Check: two outcomes (coupon or not), independent customers, n = 10, p = 0.3 for each. Binomial.
- Step 2: P(X = 3) = C(10, 3)(0.3)³(0.7)⁷ = 120 × 0.027 × 0.0823543 ≈ 0.267. (binompdf(10, 0.3, 3))
- Step 3: P(X ≤ 2) = binomcdf(10, 0.3, 2) ≈ 0.383.
- Step 4: P(X ≥ 3) = 1 − P(X ≤ 2) ≈ 1 − 0.383 = 0.617.
Answer: P(X = 3) ≈ 0.267, P(X ≤ 2) ≈ 0.383, P(X ≥ 3) ≈ 0.617.
- Example 2Calculator allowed
Is the result surprising?
Using the same store, 0 of 10 randomly chosen customers used a coupon. Is that surprising if p really is 0.3?
Show the solutionHide the solution
- Step 1: P(X = 0) = (0.7)¹⁰ ≈ 0.028.
- Step 2: That happens in about 2.8% of groups of 10 customers. It's possible but unusual.
- Step 3: It's some evidence that coupon use may be lower than 30% for these customers, an idea you'll formalize with significance tests in Unit 3.
Answer: P(X = 0) ≈ 0.028, so it's fairly surprising if p = 0.3.
- Example 3Calculator allowed
Trap: not binomial
A bag holds 6 red and 4 blue marbles. You draw 3 without replacement and let X = number of red marbles. Is X binomial?
Show the solutionHide the solution
- Step 1: Two outcomes (red or not) and a fixed n = 3 are fine.
- Step 2: But without replacement, p changes: 6/10 on the first draw, then 5/9 or 6/9 on the second, depending on the first.
- Step 3: 3 is 30% of 10, far more than 10%, so the trials aren't close to independent.
Answer: No. The probability of red changes from draw to draw, so the trials aren't independent and p isn't constant.
Common mistakes
- Using binomcdf(n, p, x) for "at least x." It gives "at most x."
- Calling a setting binomial without checking that p stays the same and trials are independent.
- Forgetting the square root in σX = √(np(1 − p)).
- Writing calculator syntax as the only work on free response. Name the distribution and its parameters too, like "X is binomial with n = 10 and p = 0.3."
On the exam
- Show work as: "X ~ binomial with n = 10, p = 0.3; P(X ≥ 3) = 1 − P(X ≤ 2) ≈ 0.617." Readers want the distribution, parameters and the event, not just the calculator command.
- "Justify that X is binomial" means addressing all four conditions in context.
Connected topics
Videos
Check yourself
4 questions on 2.10 The Binomial Distribution. Pick an answer to see if you got it, and why.
In a large city, 30% of adults have a library card. A random sample of 12 adults from the city is selected, and X is the number of them who have a library card.
Described scenario with an invented percent
Which condition for a binomial setting is most in doubt if the 12 adults all come from just three families instead of being chosen at random from the whole city?
What is the probability that exactly 4 of the 12 adults have a library card?
What are the mean and standard deviation of X?
Suppose 6 of the 12 randomly selected adults have a library card. Given that P(X ≥ 6) ≈ 0.118, what is the best conclusion?
0 of 4 answered