How do I know if a problem is binomial or geometric?

Both involve independent success/failure trials with the same probability. The key difference is the count. Binomial fixes the number of trials n in advance and counts successes. Geometric keeps going until the first success, so the number of trials is not fixed. If the problem says 'in 15 attempts' it is binomial; if it says 'until the first success' it is geometric.

When can I treat sampling without replacement as binomial?

Strictly, sampling without replacement makes trials dependent because p changes after each draw. The AP exam lets you treat the setting as approximately binomial when your sample is less than 10% of the population, known as the 10% condition. This keeps the probability change small enough to ignore.

What is the difference between binompdf and binomcdf?

binompdf(n,p,k) gives the probability of exactly k successes, P(X=k). binomcdf(n,p,k) gives the cumulative probability of k or fewer successes, P(X≤ k). Use pdf for 'exactly' and cdf for 'at most'; for 'at least' or 'more than' use the complement, 1-binomcdf.

Why is the standard deviation largest when p equals 0.5?

The variance np(1-p) is maximized when p=0.5 because the product p(1-p) peaks there. Intuitively, outcomes are most unpredictable when success and failure are equally likely; as p moves toward 0 or 1 the results become more certain, so the spread shrinks.

AP-STATS-4.9-4.10

U4.9 Binomial Distribution

Master the binomial distribution for AP Statistics: check BINS conditions, compute P(X=k), find mean and SD, and use binompdf/binomcdf on your calculator.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4.9 Binomial Distribution, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

How many free throws will a 78% shooter make in 10 attempts? How many defective chips appear in a batch of 20? Whenever you count successes in a fixed number of independent trials, you are working with a binomial distribution — one of the most tested ideas in AP Statistics Unit 4.

In this lesson you will learn to recognize a binomial setting using the BINS checklist, compute exact probabilities with the binomial formula, summarize the distribution with its mean and standard deviation, and speed everything up with the binompdf and binomcdf calculator commands. Getting comfortable here sets you up for the sampling distributions in Unit 5.

Recognizing a Binomial Setting: BINS

A random variable is binomial only when all four BINS conditions hold. Missing even one means it is not binomial.

Letter	Condition	What to check
B	Binary	Each trial has exactly two outcomes: success or failure
I	Independent	The result of one trial does not affect another
N	Number fixed	The number of trials $n$ is set in advance
S	Same probability	The success probability $p$ is identical on every trial

The binomial variable

X

counts the number of successes in those

n

trials, so

X

can take whole-number values from

0

n

.

A common misconception: students assume any yes/no scenario is binomial. But if you keep sampling until something happens (no fixed

n

), that is a geometric setting, taught in U4.11 — not binomial. Another trap is independence when sampling without replacement. Technically drawing cards without replacement changes

p

each draw. The exam allows treating it as approximately binomial when the sample is less than 10% of the population (the 10% condition). Watch for wording like 'exactly 12 trials' (fixed

n

, binomial) versus 'until the first success' (geometric).

Computing P(X = k) with the Formula

The probability of getting exactly

k

successes in

n

trials is

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

The three pieces each have meaning. The combination

\binom{n}{k}=\frac{n!}{k!(n-k)!}

counts how many different orders produce

k

successes. The term

p^k

is the probability of those

k

successes, and

(1-p)^{n-k}

is the probability of the remaining failures. Multiplying combines the arrangements with the probability of each arrangement.

For example, with

n=5

and

p=0.4

, the chance of exactly

2

successes is

\binom{5}{2}(0.4)^2(0.6)^3=10(0.16)(0.216)=0.3456

.

The binomial coefficient explains why the distribution is symmetric when

p=0.5

and skewed otherwise. When

p<0.5

the distribution is skewed right; when

p>0.5

it is skewed left. The exam often asks 'at least' or 'at most' questions, which require adding several

P(X=k)

values — this is where the calculator's cumulative command saves time. Always identify

n

p

, and exactly what count you want before plugging in.

Mean and Standard Deviation of a Binomial

Because a binomial variable is a sum of

n

independent trials, its center and spread have clean formulas:

\mu_X = np \qquad \sigma_X = \sqrt{np(1-p)}

The mean makes intuitive sense: if you shoot 10 free throws at

p=0.78

, you expect about

10(0.78)=7.8

makes. The standard deviation measures typical variation around that mean, here

\sqrt{10(0.78)(0.22)}\approx1.31

makes.

A frequent error is to use

\sqrt{np}

or to forget the

(1-p)

factor. Remember the variance is

np(1-p)

, and the SD is its square root. The spread is largest when

p=0.5

and shrinks as

p

approaches

0

1

, because outcomes become more predictable at the extremes.

The exam commonly pairs these formulas with an interpretation sentence. Be ready to write, in context, something like: 'In many groups of 10 shots, the number of makes typically varies by about 1.31 from the mean of 7.8.' Interpreting the standard deviation in context earns points that bare calculations do not.

Calculator: binompdf and binomcdf

Your calculator computes binomial probabilities directly, and you should show both the setup and the numeric answer on the exam.

Command	Computes	Use when the question says
$\text{binompdf}(n,p,k)$	$P(X=k)$	'exactly', a single value
$\text{binomcdf}(n,p,k)$	$P(X\le k)$	'at most', 'no more than', 'fewer than or equal'

For 'at least' or 'more than' questions, use the complement. For instance

P(X\ge k)=1-\text{binomcdf}(n,p,k-1)

, and

P(X>k)=1-\text{binomcdf}(n,p,k)

. Translating the words into the correct boundary is the skill graders test most.

A quick reference: 'fewer than 4' means

X\le 3

, so use

\text{binomcdf}(n,p,3)

; 'at least 4' means

X\ge 4

, so use

1-\text{binomcdf}(n,p,3)

. On free-response questions, name the command and its inputs, for example 'binomcdf(20, 0.1, 3)', rather than writing only calculator syntax with no labels. This shows the reader you chose the right distribution and boundary.

Key terms

Binomial setting.: A situation satisfying BINS: binary outcomes, independent trials, a fixed number of trials, and the same success probability on each trial.
Binomial random variable.: The count $X$ of successes in $n$ independent trials, taking whole-number values from $0$ to $n$ .
Success probability (p).: The fixed probability that any single trial results in a success; must be identical across all trials.
Binomial coefficient.: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ , the number of distinct arrangements of $k$ successes among $n$ trials.
10% condition.: When sampling without replacement, trials can be treated as approximately independent if the sample is less than 10% of the population.
binompdf.: Calculator command giving $P(X=k)$ , the probability of exactly $k$ successes.
binomcdf.: Calculator command giving $P(X\le k)$ , the cumulative probability of at most $k$ successes.

Worked example

A basketball player makes 78% of her free throws. She attempts 10 free throws, and attempts are independent. Let

X

be the number she makes. (a) Explain why

X

is binomial. (b) Find

P(X=8)

. (c) Find the probability she makes at least 8. (d) Find and interpret the mean and standard deviation.

Part (a): Check BINS. Binary — each shot is made or missed. Independent — stated. Number fixed —

n=10

attempts. Same probability —

p=0.78

each shot. All four hold, so

X

is binomial with

n=10

p=0.78

.

Part (b): Use the formula

P(X=8)=\binom{10}{8}(0.78)^8(0.22)^2

. Here

\binom{10}{8}=45

(0.78)^8\approx0.1370

, and

(0.22)^2=0.0484

. Multiplying:

45(0.1370)(0.0484)\approx0.2984

. On the calculator this is binompdf(10, 0.78, 8).

Part (c): 'At least 8' means

X\ge 8

, so

P(X\ge 8)=P(8)+P(9)+P(10)

. Using the complement:

1-\text{binomcdf}(10,0.78,7)\approx 1-0.4664=0.5336

.

Part (d): Mean

\mu=np=10(0.78)=7.8

makes. Standard deviation

\sigma=\sqrt{10(0.78)(0.22)}=\sqrt{1.716}\approx1.31

makes. Interpretation: over many sets of 10 attempts, she averages 7.8 makes, and the number of makes typically varies by about 1.31 from that average.

Practice questions

A quality inspector checks 20 randomly selected light bulbs from a large shipment in which 5% are defective. Let

X

be the number of defective bulbs. Which expression gives the probability that at most 2 are defective?

$\text{binompdf}(20, 0.05, 2)$
$\text{binomcdf}(20, 0.05, 2)$
$1-\text{binomcdf}(20, 0.05, 2)$
$1-\text{binomcdf}(20, 0.05, 1)$

Answer: $\text{binomcdf}(20, 0.05, 2)$

'At most 2' means

X\le 2

, which is exactly what the cumulative command binomcdf returns at

k=2

. binompdf would give only

P(X=2)

; the complement expressions would give 'more than' probabilities. The sample of 20 is under 10% of the large shipment, so treating the trials as binomial is justified.

A multiple-choice quiz has 8 questions, each with 4 options, and a student guesses randomly on every question. Find the mean and standard deviation of the number of correct answers, and interpret the standard deviation in context.

Answer: $\mu = 2$ correct answers and $\sigma \approx 1.22$ correct answers.

Each guess is a trial with

p=1/4=0.25

and

n=8

. The mean is

\mu=np=8(0.25)=2

correct answers. The standard deviation is

\sigma=\sqrt{np(1-p)}=\sqrt{8(0.25)(0.75)}=\sqrt{1.5}\approx1.22

. Interpretation: across many such randomly-guessed quizzes, the number of correct answers typically varies by about 1.22 from the average of 2.

A fair six-sided die is rolled 12 times. Let

X

count the number of times a 6 appears. Find

P(X=3)

Answer: About $0.197$ .

This is binomial with

n=12

p=1/6

. Using the formula,

P(X=3)=\binom{12}{3}(1/6)^3(5/6)^9=220(0.00463)(0.1938)\approx0.197

. Equivalently, binompdf(12, 1/6, 3) gives roughly 0.197. Each roll is binary (6 or not), independent, fixed at 12 trials, with constant probability

1/6

, so BINS is satisfied.

FAQ

How do I know if a problem is binomial or geometric?: Both involve independent success/failure trials with the same probability. The key difference is the count. Binomial fixes the number of trials $n$ in advance and counts successes. Geometric keeps going until the first success, so the number of trials is not fixed. If the problem says 'in 15 attempts' it is binomial; if it says 'until the first success' it is geometric.
When can I treat sampling without replacement as binomial?: Strictly, sampling without replacement makes trials dependent because $p$ changes after each draw. The AP exam lets you treat the setting as approximately binomial when your sample is less than 10% of the population, known as the 10% condition. This keeps the probability change small enough to ignore.
What is the difference between binompdf and binomcdf?: binompdf $(n,p,k)$ gives the probability of exactly $k$ successes, $P(X=k)$ . binomcdf $(n,p,k)$ gives the cumulative probability of $k$ or fewer successes, $P(X\le k)$ . Use pdf for 'exactly' and cdf for 'at most'; for 'at least' or 'more than' use the complement, $1-\text{binomcdf}$ .
Why is the standard deviation largest when p equals 0.5?: The variance $np(1-p)$ is maximized when $p=0.5$ because the product $p(1-p)$ peaks there. Intuitively, outcomes are most unpredictable when success and failure are equally likely; as $p$ moves toward 0 or 1 the results become more certain, so the spread shrinks.

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