How do I know whether to use the binomial or geometric formula on an FRQ?

Read what is being counted. If the number of trials is fixed and you count successes among them, it's binomial. If you keep going until the first success and count how many trials that takes, it's geometric. Phrases like 'exactly k of n' signal binomial; 'first success' or 'until' signal geometric.

Do I lose points for not showing the formula if my answer is correct?

Often yes. AP readers award points for communication and process, so a correct number with no supporting work is risky. Write the general formula, substitute the values, and report the result with units. This also earns partial credit if your final arithmetic slips.

Why can't I add standard deviations when combining two random variables?

Variability combines through variance, not standard deviation. For independent variables, ^2_X± Y=_X^2+_Y^2, and you take the square root at the end. Adding standard deviations directly overestimates the true spread and will be marked wrong.

What does interpreting an expected value require for full credit?

Three things: the phrase 'on average' or 'in the long run,' the quantity described in the context of the problem, and correct units. Remember an expected value can be a non-integer even for counts, and it need not be an outcome you could actually observe.

AP-STATS-4-FRQ

U4 FRQ Practice

Master AP Statistics Unit 4 free-response questions: probability rules, random variables, binomial and geometric distributions, plus scoring-ready communication strategies.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U4 FRQ Practice free

What this lesson covers

Unit 4 packs the AP exam's most formula-heavy probability content into free-response questions that reward clear setup and clean communication. Readers don't just want a number — they want to see the probability rule you used, the distribution you identified, and units on every answer. This guide shows you how to attack a Unit 4 FRQ: how to recognize which technique the prompt is fishing for, how to organize your work so a reader can follow it, and how to avoid the small omissions that cost points. You already know the individual tools from earlier lessons; here we drill the meta-skill of choosing the right one under time pressure and writing it up like a scorer wants to see.

Decoding what the FRQ is asking

The first job on any Unit 4 FRQ is diagnosis: which probability tool does the wording demand? AP prompts embed signal words that point to a specific technique. Reading these correctly before you compute prevents the most common error — answering a different question than the one asked.

Prompt language	Likely tool	What to show
"probability that... given that..."	Conditional probability	$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
"exactly $k$ of $n$ "	Binomial	$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$
"first success on trial $k$ "	Geometric	$P(X=k)=(1-p)^{k-1}p$
"expected value" or "long-run average"	Mean of a random variable	$\mu_X=\sum x_iP(x_i)$
"total" or "difference" of two variables	Combining random variables	$\mu_{X\pm Y}$ , $\sigma^2_{X\pm Y}=\sigma_X^2+\sigma_Y^2$

Circle the numbers and the exact question in each part. A part that says "interpret" wants a sentence in context, not a calculation. A part that says "justify" wants you to name the rule or condition explicitly. Matching your response type to the verb is worth as much as arithmetic accuracy.

Setting up probability and distribution work for full credit

AP readers score communication, so the structure of your answer matters. Define your random variable in words before using it: writing "Let

X

= number of defective chips in a sample of 12" earns the setup point and forces you to track units. Then state the distribution and its parameters, for example "

X

follows a binomial distribution with

n=12

and

p=0.05

."

When you use a rule, write the general formula, then the substituted version, then the result. A bare answer of

0.882

with no supporting expression is fragile — if it's wrong, there's nothing to earn partial credit. Showing

P(X\ge 1)=1-P(X=0)=1-(0.95)^{12}=0.4596

protects you.

A frequent misconception is confusing "at least one" with "exactly one." "At least one" almost always uses the complement,

1-P(X=0)

. Another trap: adding standard deviations when combining independent random variables. You add variances, never standard deviations, then take the square root:

\sigma_{X+Y}=\sqrt{\sigma_X^2+\sigma_Y^2}

. Finally, every probability must land between 0 and 1 — a quick sanity check that catches setup errors before you commit.

Interpreting results in context

Unit 4 FRQs increasingly ask you to interpret rather than just compute. An expected value is a long-run average, so an interpretation must include the phrase "on average" or "in the long run," the variable in context, and correct units. For instance: "Over many randomly selected samples of 12 chips, the store would average about 0.6 defective chips per sample."

Notice that expected value need not be a possible outcome — you cannot observe 0.6 defective chips in a single sample, and that's fine. Students often round expected values to whole numbers thinking a count must be an integer; don't. Report

\mu_X=0.6

chips.

When interpreting a standard deviation, describe typical deviation from the mean in context and units. When asked whether an outcome is "surprising" or provides "convincing evidence," compute the relevant probability and compare it to a reasonable threshold, then state a conclusion in context. Vague answers like "it's unlikely" without a probability lose the point. Tie the number to the scenario: "Because

P(X\ge 5)=0.0004

is very small, observing 5 defects would be surprising if the defect rate really were 5%."

Independence, conditions, and justification parts

Many Unit 4 FRQs contain a part asking you to check or argue independence, or to justify using a particular distribution. These are quick points if you know the exact statements readers look for.

To show two events are independent, verify

P(A\mid B)=P(A)

or, equivalently,

P(A\cap B)=P(A)\cdot P(B)

. Show the actual numbers on both sides and state whether they match. To justify a binomial model, confirm a fixed number of trials, two outcomes per trial, constant probability of success, and independent trials. For a geometric model, the same conditions hold except you count trials until the first success rather than fixing

n

.

A common misconception is treating "mutually exclusive" as "independent." They are nearly opposite: if two events with nonzero probability cannot both occur, then knowing one occurred changes the other's probability to zero, so they are dependent. When a part says "explain whether the assumption of independence is reasonable," you usually cannot prove it mathematically — instead give a contextual argument, such as whether sampling without replacement from a large population keeps trial probabilities roughly constant (the 10% condition).

Key terms

Random variable.: A variable whose numeric value is determined by the outcome of a random process; denoted with a capital letter such as $X$ .
Expected value.: The long-run average of a random variable, computed as $\mu_X=\sum x_iP(x_i)$ ; it need not be an attainable outcome.
Binomial distribution.: Models the count of successes in a fixed number $n$ of independent trials with constant success probability $p$ .
Geometric distribution.: Models the number of trials up to and including the first success, with constant success probability $p$ .
Conditional probability.: The probability of one event given another has occurred: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ .
Independence.: Two events are independent when the occurrence of one does not change the probability of the other: $P(A\mid B)=P(A)$ .
Complement rule.: For any event, $P(\text{at least one})=1-P(\text{none})$ ; central to 'at least one' problems.
Variance addition rule.: For independent random variables, $\sigma^2_{X\pm Y}=\sigma_X^2+\sigma_Y^2$ ; standard deviations are never added directly.

Worked example

A quality inspector checks memory chips one at a time. Each chip is defective independently with probability 0.05. (a) In a random sample of 12 chips, find the probability that exactly 1 is defective. (b) Find the probability that at least 1 of the 12 is defective. (c) The inspector instead tests chips one at a time until the first defective chip is found. Find the probability that the first defective chip is the 4th one tested. (d) Interpret the expected number of chips tested until the first defective in context.

Define the variables first. For (a) and (b), let

X

= number of defective chips in 12, so

X

is binomial with

n=12

and

p=0.05

.

(a) Use the binomial formula:

P(X=1)=\binom{12}{1}(0.05)^1(0.95)^{11}

. Compute

\binom{12}{1}=12

, so

P(X=1)=12(0.05)(0.5688)=0.3413

.

(b) 'At least 1' calls for the complement:

P(X\ge 1)=1-P(X=0)=1-(0.95)^{12}=1-0.5404=0.4596

.

(c) Now the process changes. Let

Y

= trial on which the first defective appears, so

Y

is geometric with

p=0.05

. Then

P(Y=4)=(0.95)^3(0.05)=0.8574\times0.05=0.0429

.

(d) The mean of a geometric distribution is

\mu_Y=\frac{1}{p}=\frac{1}{0.05}=20

. Interpretation: over many repetitions of this testing process, the inspector would test an average of about 20 chips before finding the first defective one. State it in context and note it is a long-run average, not a guaranteed value.

Practice questions

A basketball player makes each free throw independently with probability 0.8. She shoots 5 free throws. Which expression gives the probability she makes exactly 4?

$\binom{5}{4}(0.8)^4(0.2)^1$
$\binom{5}{4}(0.2)^4(0.8)^1$
$(0.8)^4(0.2)^1$
$1-(0.2)^5$

Answer: $\binom{5}{4}(0.8)^4(0.2)^1$

This is binomial with

n=5

p=0.8

, and

k=4

. The formula is

\binom{5}{4}(0.8)^4(0.2)^{5-4}

. The binomial coefficient

\binom{5}{4}=5

counts the ways to choose which 4 shots are made,

(0.8)^4

is the probability of those makes, and

(0.2)^1

covers the single miss. The last choice would answer 'at least one make,' a different question.

Two independent components have lifetimes with means

\mu_X=40

and

\mu_Y=30

months and standard deviations

\sigma_X=6

and

\sigma_Y=8

months. Find the mean and standard deviation of the total lifetime

T=X+Y

, and explain why you cannot simply add the standard deviations.

Answer: $\mu_T=70$ months and $\sigma_T=10$ months.

Means always add:

\mu_T=40+30=70

months. For independent variables, variances add:

\sigma_T^2=6^2+8^2=36+64=100

, so

\sigma_T=\sqrt{100}=10

months. You cannot add standard deviations (

6+8=14

) because variability accumulates through variance, not standard deviation; adding standard deviations would overstate the spread. The independence assumption is what allows the variances to be added with no covariance term.

A machine produces parts that are acceptable 92% of the time, independently. Parts are inspected until the first unacceptable part is found. Find the probability the first unacceptable part occurs on or before the 3rd inspection.

Answer: Approximately 0.2213.

Let

Y

= trial of the first unacceptable part, geometric with

p=0.08

(unacceptable probability). Use the complement:

P(Y\le 3)=1-P(Y>3)=1-(0.92)^3=1-0.7787=0.2213

. Alternatively sum

P(Y=1)+P(Y=2)+P(Y=3)

. Naming the success as 'unacceptable' with

p=0.08

is the key setup step; using 0.92 as

p

would answer the wrong question.

FAQ

How do I know whether to use the binomial or geometric formula on an FRQ?: Read what is being counted. If the number of trials is fixed and you count successes among them, it's binomial. If you keep going until the first success and count how many trials that takes, it's geometric. Phrases like 'exactly $k$ of $n$ ' signal binomial; 'first success' or 'until' signal geometric.
Do I lose points for not showing the formula if my answer is correct?: Often yes. AP readers award points for communication and process, so a correct number with no supporting work is risky. Write the general formula, substitute the values, and report the result with units. This also earns partial credit if your final arithmetic slips.
Why can't I add standard deviations when combining two random variables?: Variability combines through variance, not standard deviation. For independent variables, $\sigma^2_{X\pm Y}=\sigma_X^2+\sigma_Y^2$ , and you take the square root at the end. Adding standard deviations directly overestimates the true spread and will be marked wrong.
What does interpreting an expected value require for full credit?: Three things: the phrase 'on average' or 'in the long run,' the quantity described in the context of the problem, and correct units. Remember an expected value can be a non-integer even for counts, and it need not be an outcome you could actually observe.

Learn this with a teacher, not a page

The Crimsora tutor teaches U4 FRQ Practice live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free