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

Look at what is fixed. If the number of trials n is fixed and you count successes, it is binomial. If instead you keep going until the first success and count how many trials that takes, it is geometric. Phrases like "until the first" or "how many attempts until" signal geometric.

Why is the exponent k−1 instead of k in the geometric formula?

To have your first success on trial k, the previous k-1 trials must all be failures, and then trial k is the success. So you multiply (1-p) by itself k-1 times for the failures and multiply by p once for the success, giving (1-p)^k-1p.

When should I use geomcdf instead of geompdf?

Use geompdf(p,k) for an exact single value, P(X=k). Use geomcdf(p,k) for a cumulative range, P(X ≤ k), which answers "at most k trials" or "within the first k trials." For "more than k," compute 1 - geomcdf(p,k).

Can a geometric random variable take an infinite number of values?

Yes. In theory X can be any positive integer 1, 2, 3, with no upper bound, since it is always possible (though increasingly unlikely) to keep failing. The probabilities shrink geometrically and sum to 1, and the expected value 1/p is finite.

AP-STATS-4.11

U4.11 Geometric Distribution

Master the AP Statistics geometric distribution: check BITS conditions, compute P(X=k), find the mean 1/p and standard deviation, and use geompdf/geomcdf.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4.11 Geometric Distribution, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

In Unit 4.9 you learned to count successes in a fixed number of trials with the binomial distribution. The geometric distribution flips the question around: instead of asking "how many successes in

n

trials," it asks "how many trials until the first success?" Think of shooting free throws until you finally make one, or rolling a die until the first six appears.

This lesson shows you how to recognize a geometric setting using the BITS conditions, compute the probability that the first success happens on trial

k

, and find the mean and standard deviation of the waiting time. You will also learn the calculator shortcuts geompdf and geomcdf, which the exam expects you to use fluently.

Recognizing a Geometric Setting: BITS

A random variable

X

follows a geometric distribution when the process meets four conditions, remembered by the acronym BITS.

B — Binary: each trial results in one of two outcomes, labeled success or failure.

I — Independent: the outcome of one trial does not affect any other. When sampling without replacement, the 10% condition (sample under 10% of the population) keeps trials approximately independent.

T — Trials until success: you keep going until the first success occurs. This is the defining feature — the number of trials is not fixed.

S — Same probability: the probability of success

p

is identical on every trial.

The key contrast with the binomial is the T. In a binomial setting you fix the number of trials

n

and count successes. In a geometric setting the number of successes is fixed at exactly one (the first), and you count the trials needed to get there. The random variable

X

is the trial number on which the first success occurs, so

X

can be

1, 2, 3, \dots

with no upper limit.

A common exam trap: a problem that says "how many attempts until" signals geometric, while "how many successes in a fixed number" signals binomial. Read carefully to spot which quantity is fixed.

Computing P(X = k)

To have the first success on trial

k

, you must fail the first

k-1

trials and then succeed on trial

k

. Because trials are independent with constant probability, you multiply:

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

Here

p

is the probability of success on a single trial and

1-p

is the probability of failure. The exponent is

k-1

, not

k

— this is the most frequent arithmetic mistake students make. You fail

k-1

times, then succeed once.

For example, if

p = 0.2

, then

P(X = 4) = (0.8)^{3}(0.2) = 0.512 \times 0.2 = 0.1024

.

Sometimes a question asks for the probability that the first success takes more than

k

trials. This means the first

k

trials were all failures:

P(X > k) = (1-p)^{k}

This shortcut is handy because it avoids summing an infinite series. From it you can also get

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

, which is exactly what the cumulative calculator command returns. Knowing these algebraic forms lets you check calculator output and answer conceptual questions without a device.

Mean and Standard Deviation

On average, how many trials until the first success? Intuitively, if success happens 1 in every

p

fraction of trials, you expect to wait

1/p

trials. That is exactly the mean:

\mu_X = \frac{1}{p}

The standard deviation is:

\sigma_X = \sqrt{\frac{1-p}{p^{2}}} = \frac{\sqrt{1-p}}{p}

For instance, if

p = 0.25

, then

\mu = 1/0.25 = 4

trials on average, and

\sigma = \sqrt{0.75}/0.25 = 0.866/0.25 \approx 3.46

trials.

Notice the mean makes sense: a rare event (small

p

) means a long expected wait. If

p = 0.1

, you expect 10 trials. Interpretation matters on the exam — write "we expect about 4 trials until the first success," not just "

\mu = 4

Feature	Geometric	Binomial
Question	Trials until 1st success	Successes in $n$ trials
Fixed	Number of successes (1)	Number of trials $n$
Mean	$\frac{1}{p}$	$np$
SD	$\frac{\sqrt{1-p}}{p}$	$\sqrt{np(1-p)}$

Memorizing this comparison table prevents mixing up the two formulas under time pressure.

Using geompdf and geomcdf

Your calculator has two geometric functions found in the distribution menu.

geompdf(p, k) gives the probability of the first success on exactly trial

k

, i.e.

P(X = k)

. The "pdf" stands for probability density function — use it for an exact single value.

geomcdf(p, k) gives the cumulative probability

P(X \le k)

, the chance the first success happens on or before trial

k

. The "cdf" is cumulative — use it for "at most" or "within the first

k

trials."

You want	Command
$P(X = k)$	geompdf $(p,k)$
$P(X \le k)$	$\text{geomcdf}(p,k)$
$P(X < k)$	$\text{geomcdf}(p,k-1)$
$P(X \ge k)$	$1 - \text{geomcdf}(p,k-1)$
$P(X > k)$	$1 - \text{geomcdf}(p,k)$

Order of arguments matters: probability

p

comes first, then the trial number

k

. On free-response questions, always show the setup — for example write "

P(X=4) =

geompdf

(0.2, 4) = 0.1024

" — because naked calculator syntax without defining the variable or parameters may not earn full communication credit.

Key terms

Geometric random variable.: A count of the number of independent trials needed to obtain the first success, where each trial has the same success probability $p$ .
BITS conditions.: The four requirements for a geometric setting: Binary outcomes, Independent trials, Trials until first success, and Same probability on each trial.
Success probability (p).: The constant probability that any single trial results in a success; drives the mean, SD, and every probability calculation.
P(X = k).: The probability the first success occurs on trial $k$ , equal to $(1-p)^{k-1}p$ : fail $k-1$ times then succeed.
geompdf.: Calculator command returning $P(X=k)$ for a geometric variable, taking arguments $(p, k)$ .
geomcdf.: Calculator command returning the cumulative probability $P(X \le k)$ for a geometric variable, taking arguments $(p, k)$ .
Mean of a geometric distribution.: The expected number of trials until the first success, $\mu = 1/p$ .
10% condition.: When sampling without replacement, keeping the sample under 10% of the population so trials stay approximately independent.

Worked example

A basketball player makes each free throw independently with probability

p = 0.7

. Let

X

be the number of free throws until (and including) her first miss. (a) Explain why

X

is geometric. (b) Find the probability her first miss is on the third attempt. (c) Find the mean and standard deviation of

X

. (d) Find the probability her first miss occurs within the first four attempts.

First, define the "success" carefully. Since

X

counts trials until the first miss, a "success" here is a MISS, with probability

p = 1 - 0.7 = 0.3

. This relabeling is a classic trick — the event you are waiting for becomes the success.

(a) Check BITS. Binary: each shot is made or missed. Independent: shots are stated independent. Trials until success: we count shots until the first miss. Same probability:

p = 0.3

each shot. All four hold, so

X

is geometric with

p = 0.3

.

(b) First miss on the third attempt means make, make, miss:

P(X=3) = (1-0.3)^{2}(0.3) = (0.7)^2(0.3) = 0.49 \times 0.3 = 0.147

. On the calculator, geompdf

(0.3, 3) = 0.147

.

(c) Mean:

\mu = 1/p = 1/0.3 \approx 3.33

shots. Standard deviation:

\sigma = \sqrt{(1-0.3)/0.3^2} = \sqrt{0.7/0.09} = \sqrt{7.78} \approx 2.79

shots. Interpret: we expect about 3.33 shots until her first miss.

(d) "Within the first four attempts" means

P(X \le 4) =

\text{geomcdf}(0.3, 4)

. Compute:

1 - (0.7)^4 = 1 - 0.2401 = 0.7599

. So there is about a 76.0% chance her first miss happens by the fourth shot.

Practice questions

A quality inspector tests light bulbs one at a time from a large shipment. Each bulb is defective with probability 0.05, independent of others. Let

X

be the number of bulbs tested until the first defective bulb is found. What is the probability that the first defective bulb is the sixth one tested?

$(0.05)^{5}(0.95)$
$(0.95)^{5}(0.05)$
$(0.95)^{6}(0.05)$
$(0.05)^{6}(0.95)$

Answer: $(0.95)^{5}(0.05)$

The first success (a defective bulb) occurs on trial

k = 6

with

p = 0.05

. Using

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

, we fail 5 times (non-defective, probability 0.95 each) then succeed once:

(0.95)^{5}(0.05)

. The exponent is

k-1 = 5

, not 6 — a common error. This equals about 0.0387.

A student guesses randomly on multiple-choice questions, each with 5 options, so the probability of a correct guess is 0.2. Let

X

be the number of questions until the first correct guess. Find and interpret the mean of

X

, and find the probability it takes more than 5 questions to get the first correct answer. Show your work.

Answer: Mean $\mu = 5$ questions; $P(X > 5) = (0.8)^5 \approx 0.328$ .

The setting is geometric with

p = 0.2

. The mean is

\mu = 1/p = 1/0.2 = 5

, interpreted as: on average the student needs about 5 questions to get the first correct guess. For

P(X > 5)

, the first 5 guesses must all be wrong:

P(X > 5) = (1-p)^{5} = (0.8)^{5} = 0.32768

. On a calculator this equals

1 - \text{geomcdf}(0.2, 5)

. So there is about a 32.8% chance more than 5 questions are needed.

Which of the following scenarios describes a geometric random variable rather than a binomial one?

The number of heads in 20 coin flips
The number of defective items in a batch of 50
The number of cards drawn until the first ace appears
The proportion of voters supporting a candidate in a sample of 100

Answer: The number of cards drawn until the first ace appears

A geometric variable counts trials until the first success, with the number of trials not fixed. Drawing cards until the first ace fits this. The other options have a fixed number of trials (20 flips, batch of 50, sample of 100) and count successes or a proportion, which are binomial or proportion settings.

FAQ

How do I know whether a problem is geometric or binomial?: Look at what is fixed. If the number of trials $n$ is fixed and you count successes, it is binomial. If instead you keep going until the first success and count how many trials that takes, it is geometric. Phrases like "until the first" or "how many attempts until" signal geometric.
Why is the exponent k−1 instead of k in the geometric formula?: To have your first success on trial $k$ , the previous $k-1$ trials must all be failures, and then trial $k$ is the success. So you multiply $(1-p)$ by itself $k-1$ times for the failures and multiply by $p$ once for the success, giving $(1-p)^{k-1}p$ .
When should I use geomcdf instead of geompdf?: Use geompdf $(p,k)$ for an exact single value, $P(X=k)$ . Use $\text{geomcdf}(p,k)$ for a cumulative range, $P(X \le k)$ , which answers "at most $k$ trials" or "within the first $k$ trials." For "more than $k$ ," compute $1 - \text{geomcdf}(p,k)$ .
Can a geometric random variable take an infinite number of values?: Yes. In theory $X$ can be any positive integer $1, 2, 3, \dots$ with no upper bound, since it is always possible (though increasingly unlikely) to keep failing. The probabilities shrink geometrically and sum to 1, and the expected value $1/p$ is finite.

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