Geometric Distribution Explained: Waiting for the First Success, Formula and Examples

In 30 seconds

The geometric distribution counts the number of trials up to and including the first success, when each trial succeeds with the same probability $p$.
Its probabilities are $P(X = r) = (1-p)^{r-1}p$ for $r = 1, 2, 3, \dots$, so every bar is a fixed fraction $1 – p$ of the one before.
The mean is $1/p$ and the standard deviation is $\sqrt{1-p}/p$. The most likely value is always $X = 1$.
The case answer: a volunteer whose calls succeed 15% of the time gets a first donation within 5 calls with probability 0.556, waits 6.67 calls on average, and needs 15 calls to be 90% sure of a donation.
Exam point: use $P(X > r) = (1-p)^r$ for “more than” and “at least” questions. It saves time and avoids long sums.

The case: how many calls until the first donation?

Hana volunteers at a charity’s phone bank on Saturday mornings. She rings past supporters and asks for a small gift. From last year’s records, about 15% of calls end in a donation. (The figures are illustrative.) Calls are to different households, so the result of one call does not affect the next.

The team leader likes every new volunteer to “get one on the board” early, because a first success keeps morale up. Hana has three questions before she starts:

What is the chance that her first donation comes on exactly her third call?
What is the chance she gets at least one donation within her first five calls?
On average, how many calls will it take, and how many calls should she plan for if she wants to be 90% sure of a first donation?

These are all “waiting” questions. The number of successes in a fixed number of calls would be a binomial question. Here the number of calls is the random quantity, and we stop at the first success. That is the geometric distribution.

The concept: the geometric distribution

Think of a sequence of identical, independent trials. Each trial ends in success or failure. The chance of success is the same, $p$, every time. We keep going until the first success and record $X$, the number of trials it took, counting the successful one.^[1]

For the first success to happen on trial $r$, two things must happen. The first $r – 1$ trials must all fail, and trial $r$ must succeed. Because the trials are independent, we multiply the probabilities: $r – 1$ factors of $(1-p)$ and one factor of $p$.

That simple argument gives the whole distribution. There is no combination term like the $\binom{n}{r}$ in the binomial, because there is only one way for the first success to land on trial $r$: every earlier trial fails.

Key termGeometric distribution: the distribution of $X$, the number of independent trials needed to get the first success, when each trial has the same success probability $p$. We write $X \sim \text{Geo}(p)$. The possible values are $1, 2, 3, \dots$ with no upper limit.

\[ P(X = r) = (1-p)^{r-1}\,p, \qquad P(X > r) = (1-p)^{r}, \qquad E(X) = \frac{1}{p}, \qquad \operatorname{Var}(X) = \frac{1-p}{p^2} \]

Here $p$ is the probability of success on each trial and $1 – p$ is the probability of failure. $r$ is a positive whole number: the trial on which the first success occurs. $P(X > r)$ is the chance that the first $r$ trials all fail, which is why it has such a short form. Its complement gives the cumulative probability $P(X \le r) = 1 – (1-p)^r$. $E(X)$ is the long-run average number of trials, and $\operatorname{Var}(X)$ is the variance; the standard deviation is its square root, $\sqrt{1-p}/p$. The probabilities form a geometric sequence with first term $p$ and common ratio $1-p$, which is where the name comes from.

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Why this method fits the case

Check the four conditions that make a geometric model valid. A handy memory aid used in AP Statistics courses is “BITS”: Binary, Independent, Trials until success, Same probability.^[2]

Binary: each call either ends in a donation or it does not.
Independent: different households decide separately, so one result does not change the next.
Trials until success: we count calls until the first donation, not donations in a fixed number of calls.
Same probability: we assume each call has about a 15% chance. This is the condition most worth questioning.

When is the geometric model not the right tool?

When the number of trials is fixed. “How many donations in 20 calls?” is binomial, not geometric.
When you wait for the second, third or $k$th success. That needs the negative binomial distribution, which generalises the geometric.
When $p$ changes from trial to trial. If Hana gets better with practice, or if the list is sorted with the most generous supporters first, the success chance is not constant and the formulas are only rough guides.
When sampling without replacement from a small group. Drawing cards from a deck until the first ace changes the odds after each draw, so the trials are not independent.

Setting it up

Define the variable in words first, then list the numbers.

Quantity	Symbol	Value
Number of calls until the first donation	$X$	$X \sim \text{Geo}(0.15)$
Probability of success on each call	$p$	0.15
Probability of failure on each call	$1 – p$	0.85
Mean number of calls	$1/p$	$1/0.15 \approx 6.67$
Variance	$(1-p)/p^2$	$0.85/0.0225 \approx 37.78$
Standard deviation	$\sqrt{1-p}/p$	$\approx 6.15$

Notice that the standard deviation is almost as large as the mean. That is typical of waiting times: most waits are short, but a few are very long, and those long waits stretch the spread.

Solving it

Worked example

For Hana, $X \sim \text{Geo}(0.15)$. Find (a) $P(X = 3)$, (b) $P(X \le 5)$, (c) $P(X > 10)$.

Step 1(a) Two failures then a success: $P(X = 3) = 0.85^2 \times 0.15 = 0.7225 \times 0.15 = 0.1084$.

Step 2(b) Use the complement. “No donation in the first 5 calls” has probability $0.85^5 \approx 0.4437$.

Step 3So $P(X \le 5) = 1 – 0.4437 = 0.5563$.

Step 4(c) “More than 10 calls” means the first 10 all fail: $P(X > 10) = 0.85^{10} \approx 0.1969$.

Hana has a 10.8% chance of a first donation on call 3, a 55.6% chance within 5 calls, and about a 1-in-5 chance of needing more than 10 calls.

The second example turns a probability target into a number of trials. This is a very common exam question and needs logarithms.

Worked example

How many calls should Hana plan for so that she is at least 90% sure of getting her first donation? Also state the mean and standard deviation of $X$.

Step 1We need the smallest $n$ with $P(X \le n) \ge 0.9$, i.e. $1 – 0.85^n \ge 0.9$, so $0.85^n \le 0.1$.

Step 2Take logs: $n \ln 0.85 \le \ln 0.1$. Because $\ln 0.85$ is negative, the inequality flips: $n \ge \dfrac{\ln 0.1}{\ln 0.85} \approx 14.17$.

Step 3So $n = 15$. Check: $1 – 0.85^{14} \approx 0.897$ (not enough) and $1 – 0.85^{15} \approx 0.913$ (enough).

Step 4Mean $= 1/0.15 \approx 6.67$ calls; SD $= \sqrt{0.85}/0.15 \approx 6.15$ calls.

Hana should allow 15 calls to be 90% sure, more than twice the average wait of 6.67.

The third example shows the most surprising property of the geometric distribution: it has no memory.

Worked example

A factory finds that 4% of items from a machine are faulty, independently of each other. An inspector checks items one by one until she finds a faulty one. (The figures are illustrative.) Find (a) the probability the first faulty item is the 10th checked, (b) the probability more than 30 items are checked, (c) the probability more than 50 items are checked, given that the first 20 were all fine.

Step 1$X \sim \text{Geo}(0.04)$. (a) $P(X = 10) = 0.96^9 \times 0.04 \approx 0.0277$.

Step 2(b) $P(X > 30) = 0.96^{30} \approx 0.2939$.

Step 3(c) $P(X > 50 \mid X > 20) = \dfrac{P(X > 50)}{P(X > 20)} = \dfrac{0.96^{50}}{0.96^{20}} = 0.96^{30} \approx 0.2939$.

Step 4The answer to (c) equals the answer to (b). Twenty good items in a row do not make a fault “due”.

The process starts afresh after every failure: this is the memoryless property.

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What the answer tells us

Hana’s first donation is more likely than not to arrive within five calls (55.6%). Her single most likely outcome is a donation on the very first call, at 15%. Yet the average wait is 6.67 calls, and she needs 15 calls to be 90% safe. Those numbers fit together once you see the shape of the distribution in Figure 1: lots of probability on short waits, and a long tail of occasional long waits that pulls the mean to the right.

Geometric distribution, p = 0.15 P(X = r)

Probability first donation is on call rMean, 1/p

Figure 1. Each bar is 85% of the height of the bar before it. The tallest bar is always at r = 1, and the mean (6.67) sits well to the right of the most likely value. Illustrative figures.

Reasonableness check. If 15 calls in every 100 succeed, then on average there is one success per $100/15 \approx 6.67$ calls. That matches $E(X) = 1/p$ and gives a quick way to remember the mean. A second check: the probabilities must sum to 1. They form a geometric series with first term 0.15 and ratio 0.85, whose sum is $0.15/(1 – 0.85) = 1$. Good.

Sensitivity. The answers depend strongly on $p$. If a better script lifted the success rate to 20%, the mean wait would fall to 5 calls and the chance of a donation within 5 calls would rise to 0.672. If the list were older and the rate fell to 10%, the mean would rise to 10 calls and the 5-call chance would drop to 0.410. For rare events the effect is dramatic: with $p = 0.02$ the mean wait is 50 trials and the SD is about 49.5, so waits of 100 or more are quite normal. This is why a charity, a sales team or a quality-control department should estimate $p$ carefully before setting targets based on a geometric model.^[3]

Common mistakeUsing the wrong power. $P(X = r)$ has $r – 1$ failures, so the power is $r – 1$, not $r$. In contrast, $P(X > r)$ means the first $r$ trials fail, so its power is $r$. Write the sequence out in words (“fail, fail, succeed”) when in doubt.

Common mistakeBelieving a success is “due”. After a run of failures many people feel the next success must be close. The memoryless property says otherwise: the chance on the next trial is still $p$, and the expected remaining wait is still $1/p$. This is the gambler’s fallacy in a formula.

Exam tipRead the definition of $X$ in the question. Most school courses, including AP Statistics and Cambridge AS & A Level Mathematics, define $X$ as the number of trials up to and including the first success, so $X$ starts at 1 and $E(X) = 1/p$.^[4]^[5] Some university texts count failures before the first success instead, which starts at 0 and has mean $(1-p)/p$. For “at least $n$ trials” questions, remember $P(X \ge n) = P(X > n – 1) = (1-p)^{n-1}$. Our statistics guide places this topic alongside the binomial distribution.

For a fuller treatment, the Statistics Made Simple textbook covers the geometric and binomial distributions side by side, with tables showing when each applies. To build speed, the Statistics Made Simple Practice Questions book includes graded geometric problems, including the logarithm-based “how many trials” type.

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Practice questions

Give answers to 4 decimal places unless told otherwise. Open each worked answer only after you have tried the question.

EasyQ1. A fair die is rolled until a six appears. Find the probability that the first six appears on the 4th roll. Show worked answer

$X \sim \text{Geo}(1/6)$.
$P(X = 4) = (5/6)^3 \times (1/6) = 125/1296 \approx 0.0965$.

EasyQ2. A striker scores from 20% of penalties in a practice drill (the figures are illustrative), independently. Find the mean number of attempts until her first goal, and the probability she scores within her first 5 attempts. Show worked answer

$X \sim \text{Geo}(0.2)$. Mean $= 1/0.2 = 5$ attempts.
$P(X \le 5) = 1 – 0.8^5 = 1 – 0.32768 = 0.6723$.

MediumQ3. $X \sim \text{Geo}(0.25)$. Find $P(X > 5)$, $P(X = 4)$ and $P(3 \le X \le 5)$. Show worked answer

$P(X > 5) = 0.75^5 \approx 0.2373$.
$P(X = 4) = 0.75^3 \times 0.25 \approx 0.1055$.
$P(3 \le X \le 5) = P(X > 2) – P(X > 5) = 0.75^2 – 0.75^5 = 0.5625 – 0.2373 = 0.3252$.

MediumQ4. Each scratch card has a 10% chance of a prize, independently. Find the smallest number of cards, $n$, that must be bought for the probability of at least one prize among them to be at least 0.95. Show worked answer

Need $1 – 0.9^n \ge 0.95$, so $0.9^n \le 0.05$.
$n \ge \ln 0.05 / \ln 0.9 \approx 28.43$, so $n = 29$.
Check: $1 – 0.9^{28} \approx 0.9477$; $1 – 0.9^{29} \approx 0.9529$.

ExamQ5. Amir and Bina take turns to roll a fair die, Amir first. The first person to roll a six wins. (a) Find the probability that the game ends on the 4th roll. [2] (b) Show that the probability that Amir wins is $\tfrac{6}{11}$. [4] Show worked answer

(a) Three non-sixes then a six: $(5/6)^3 \times (1/6)$ [M1] $= 125/1296 \approx 0.0965$ [A1].
(b) Amir wins on roll 1, 3, 5, … : $P = \tfrac16 + (\tfrac56)^2\tfrac16 + (\tfrac56)^4\tfrac16 + \dots$ [M1 for at least two correct terms].
Geometric series with $a = \tfrac16$, ratio $(\tfrac56)^2 = \tfrac{25}{36}$ [B1].
Sum $= \dfrac{1/6}{1 – 25/36} = \dfrac{1/6}{11/36}$ [M1] $= \dfrac{6}{11}$ [A1].

Frequently asked questions

What is the difference between the geometric and binomial distributions?

Both use independent trials with a constant success probability. The binomial fixes the number of trials, $n$, and counts the successes. The geometric fixes the number of successes at one and counts the trials needed. If the question says “in 20 attempts”, think binomial; if it says “until the first”, think geometric. Our binomial distribution guide covers the other side.

Why is the most likely value always 1?

Each probability is the previous one multiplied by $1 – p$, which is less than 1. So the probabilities fall from the start, and the first one, $p$, is the largest. That is true even when $p$ is tiny. With $p = 0.02$, the single most likely trial for the first success is still trial 1, even though the average wait is 50 trials.

What does “memoryless” mean in practice?

It means past failures tell you nothing about the future. If you have already waited $k$ trials, the extra number of trials still needed has the same geometric distribution as at the start. The geometric is the only discrete distribution with this property. Its continuous cousin, the exponential distribution, has it too.

Can the geometric distribution be used for real data?

Yes, whenever the conditions are close to true: the number of attempts to pass a test, the number of customers until one buys an add-on, or the number of items until a defect. Check that the success probability is roughly stable and that trials are independent. If $p$ drifts, the model is only a first approximation.

References

Illowsky, B. and Dean, S. (2023) Introductory Statistics 2e. Houston, TX: OpenStax. https://openstax.org/details/books/introductory-statistics-2e
Starnes, D. S. and Tabor, J. The Practice of Statistics. New York: W. H. Freeman.
NIST/SEMATECH. e-Handbook of Statistical Methods. National Institute of Standards and Technology. https://www.itl.nist.gov/div898/handbook/
Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
College Board. AP Statistics Course and Exam Description. New York: College Board.

Geometric Distribution Explained: Waiting for the First Success, Formula and Examples

The case: how many calls until the first donation?

The concept: the geometric distribution

Why this method fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

What is the difference between the geometric and binomial distributions?

Why is the most likely value always 1?

What does “memoryless” mean in practice?

Can the geometric distribution be used for real data?

References

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