Geometric Distribution Calculator
What Is the Geometric Distribution?
The geometric distribution models situations where you repeat an experiment until the first success occurs. Each repetition is a Bernoulli trial — an independent experiment with exactly two outcomes: success (with probability p) and failure (with probability q = 1 − p). The distribution answers the question: how many trials will it take to get the first success?
Classic applications include the number of sales calls until a deal closes, the number of manufacturing inspections until a defective item is found, the number of free throws attempted until the first miss, or the number of website visits before a conversion. Any repeatable process with a binary outcome and constant success probability fits the geometric model.
Two conventions exist: Some textbooks define X as the trial number of the first success (X ≥ 1). Others define Y as the number of failures before the first success (Y ≥ 0). This page uses X = trial of first success by default. The relationship is Y = X − 1. The table, calculator, and formulas below clearly label which convention applies at every step.
Parameter Definitions
| Symbol | Name | Range | Description |
|---|---|---|---|
| p | Probability of success | 0 < p ≤ 1 | Constant probability of success on each independent trial |
| q | Probability of failure | q = 1 − p | Probability of failure on each trial |
| X | Random variable (trials) | x = 1, 2, 3, … | Trial number on which the first success occurs Convention 1 |
| Y | Random variable (failures) | y = 0, 1, 2, … | Failures before first success; Y = X − 1 Convention 2 |
| E(X) | Mean / Expected value | 1/p | Expected trial number of first success (Convention 1) |
| Var(X) | Variance | (1−p)/p² | Spread of the distribution; same under both conventions |
Geometric Distribution Formula
Before using the table, you must know which formula your textbook or software uses. The two conventions produce different numerical answers for the same scenario, though they describe the same underlying probability.
Convention 1 — Trial Number
X = trial number on which first success occurs. X can be 1, 2, 3, … Used by most introductory statistics textbooks, AP Statistics, and OpenStax.
Convention 2 — Failure Count
Y = number of failures before first success. Y can be 0, 1, 2, … Used by Python's SciPy (scipy.stats.geom with loc=−1), R's dgeom(), and some engineering texts.
Why the Convention Difference Matters
If a student computes P(X = 3) with p = 0.30 using Convention 1, the answer is (0.70)² × 0.30 = 0.147. Using Convention 2 for the same x = 3 gives (0.70)³ × 0.30 = 0.103 — a different number. Always check the definition used in your course materials before looking up table values or comparing software output.
Geometric Distribution Table (PMF and CDF)
Both tables below use Convention 1: X = trial number of first success (x = 1, 2, 3, …). Columns show the success probability p. Rows show the trial number x. Click any cell to load it into the calculator above.
Convention 1: X = trial number of first success. PMF: P(X = x) = (1−p)x−1 × p. CDF: P(X ≤ x) = 1 − (1−p)x. Values rounded to 4 decimal places. Click any cell to load it into the calculator.
How to Use the Geometric Distribution Table
Follow these six steps every time you use the table. The most common error is reading the wrong column — double-check that the p column matches your scenario before reading the cell value.
Step 1 — Verify the conditions
Three conditions must hold. First: each trial has exactly two outcomes (success or failure). Second: trials are independent — the result of one trial does not affect another. Third: p is constant across all trials. If any condition fails, the geometric model does not apply.
Step 2 — Identify p (success probability)
State the probability of success on a single trial. For example: a free throw shooter makes 70% of attempts → p = 0.70. The table covers p values of 0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, and 0.90. For other p values, use the calculator above.
Step 3 — Define X and choose the right convention
Decide whether X represents the trial number of the first success (Convention 1, minimum value 1) or the number of failures before the first success (Convention 2, minimum value 0). The tables on this page always use Convention 1.
Step 4 — Select PMF or CDF
Use the PMF tab for exactly x trials: P(X = x). Use the CDF tab for at most x trials: P(X ≤ x). For more than x trials: P(X > x) = 1 − P(X ≤ x). For at least x trials: P(X ≥ x) = 1 − P(X ≤ x − 1).
P(X ≤ x) → CDF table
P(X > x) = 1 − P(X ≤ x)
P(X ≥ x) = 1 − P(X ≤ x − 1)
Step 5 — Locate the cell
Find the column for your p value and the row for your x value. The intersection gives the probability. If p falls between two table columns, use the calculator above (it accepts any p between 0 and 1).
Step 6 — Interpret the result
State your answer in context. A PMF value of 0.1470 at p = 0.30, x = 3 means: there is a 14.70% probability that the first success occurs exactly on the 3rd trial. A CDF value of 0.6570 at p = 0.30, x = 3 means: there is a 65.70% probability that the first success occurs on or before the 3rd trial.
Worked Examples
Each example below defines the random variable before calculating. This step is required — skipping it is the most frequent cause of errors when using a geometric distribution table.
Example 1 — Sales Calls
Scenario: A sales representative closes a deal on any given call with probability p = 0.20. What is the probability the first closed deal occurs on the 4th call?
| Step | Action | Result |
|---|---|---|
| 1 | Define X | X = call number of first closed deal |
| 2 | State parameters | p = 0.20, q = 0.80, x = 4 |
| 3 | Write formula | P(X = 4) = (0.80)³ × 0.20 |
| 4 | Calculate | 0.512 × 0.20 = 0.1024 |
| 5 | Interpret | 10.24% probability the first deal closes on the 4th call |
Example 2 — Quality Control (CDF)
Scenario: A factory inspector finds a defective item on any inspection with probability p = 0.15. What is the probability that the first defective item is found within the first 5 inspections?
| Step | Action | Result |
|---|---|---|
| 1 | Define X | X = inspection number of first defect found |
| 2 | State parameters | p = 0.15, x = 5, need P(X ≤ 5) |
| 3 | Write formula | P(X ≤ 5) = 1 − (0.85)⁵ |
| 4 | Calculate | 1 − 0.4437 = 0.5563 |
| 5 | Interpret | 55.63% chance the first defect is found within 5 inspections |
Example 3 — Website Conversion (Complementary)
Scenario: A website converts 10% of visitors. What is the probability that more than 8 visitors arrive before the first conversion?
This asks P(X > 8) where X is the visitor number of the first conversion (Convention 1) and p = 0.10.
There is a 43.05% chance that more than 8 visitors arrive before the first conversion. Alternatively, look up the CDF table at p = 0.10, x = 8: the table shows 0.5695, and 1 − 0.5695 = 0.4305.
Mean, Variance, and Key Properties
The geometric distribution has closed-form expressions for all its properties. The mean and variance depend only on p.
The Memoryless Property
The geometric distribution is the only discrete distribution with the memoryless property. If you have already made s unsuccessful attempts, the probability of needing t more attempts is exactly the same as if you were starting fresh. Past failures carry no information about future success. This property can be counterintuitive but follows directly from the independence assumption of Bernoulli trials. It is the discrete counterpart of the memoryless property of the exponential distribution.
Geometric vs Binomial vs Negative Binomial
Three discrete distributions frequently appear in problems involving independent Bernoulli trials. Choosing the correct one depends on what you are counting and whether the number of trials is fixed.
| Feature | Geometric | Binomial | Negative Binomial |
|---|---|---|---|
| What is counted? | Trials until first success | Successes in n trials | Trials until r-th success |
| Fixed number of trials? | No | Yes (n) | No |
| Parameters | p | n, p | r, p |
| Special case of | Negative Binomial (r=1) | — | — |
| Mean | 1/p | np | r/p |
| Typical question | When will the first success occur? | How many successes in n trials? | When will the r-th success occur? |
Geometric as a Special Case of the Negative Binomial
The negative binomial distribution generalizes the geometric distribution by counting the number of trials needed to achieve r successes, where r ≥ 1. When r = 1, the negative binomial formula reduces exactly to the geometric PMF. This relationship explains why statistical software such as R's dnbinom() and Python's scipy.stats.nbinom can compute geometric probabilities by setting r = 1.
Common Mistakes and How to Avoid Them
These errors appear regularly in homework, exams, and software output. Each has a clear fix.
Mistake 1 — Using the wrong convention
Fix: Before any calculation, write out explicitly whether X counts trials (Convention 1, starts at 1) or failures (Convention 2, starts at 0). If your software uses Convention 2 (R's dgeom, Python's scipy.stats.geom), verify the output matches your definition of X.
Mistake 2 — Confusing PMF and CDF
Fix: P(X = x) is the PMF (probability of exactly x). P(X ≤ x) is the CDF (probability of x or fewer). Read the problem carefully: "exactly on the 5th trial" → PMF; "within the first 5 trials" → CDF.
Mistake 3 — Forgetting q = 1 − p in the exponent
Fix: The PMF raises q (not p) to the power x − 1. The CDF raises q to the power x. Students sometimes compute px−1 × q instead of qx−1 × p — this produces the wrong answer. Always confirm: x − 1 failures occur first, then one success.
Mistake 4 — Applying geometric when trials are not independent
Fix: If drawing without replacement from a finite population, the probability p changes each trial. Use the hypergeometric distribution instead. The geometric distribution requires independence and constant p.
Mistake 5 — Confusing geometric with binomial
Fix: Ask two questions. Is the number of trials fixed? → Binomial. Are you waiting for the first success? → Geometric. If the problem says "3 out of 10 calls result in a sale," that is binomial. If it says "how many calls until the first sale," that is geometric.
Software Implementation
Every major statistics package can compute geometric probabilities. Check the parameterization each uses before comparing results.
Excel / Google Sheets
No GEOM.DIST function. Use these formulas (Convention 1):
CDF: =1-(1-p)^x
or =NEGBINOM.DIST(x-1,1,p,0)
R
R uses Convention 2 by default (Y = failures before first success):
CDF: pgeom(x-1, p)
# x-1 converts to Convention 1
Python (SciPy)
scipy.stats.geom uses Convention 1 (X = trials):
PMF: geom.pmf(x, p)
CDF: geom.cdf(x, p)
# x = trial number (starts at 1)
Parameterization Warning
R's dgeom(x, p) treats x as the number of failures (Convention 2). To get P(X = 3) in Convention 1 notation using R, call dgeom(2, p). Python's scipy.stats.geom.pmf(x, p) uses Convention 1 directly, so geom.pmf(3, 0.30) gives P(X = 3). Always verify a known value against the formula before trusting software output for a new parameterization.
Frequently Asked Questions
What is a geometric distribution table?
A geometric distribution table lists pre-computed PMF and CDF values for the geometric distribution at standard success probabilities p and trial numbers x. It saves the effort of computing (1 − p)x−1 × p by hand.
What are the assumptions of the geometric distribution?
Three assumptions: (1) each trial has exactly two outcomes (success or failure); (2) trials are independent; (3) the probability of success p is the same on every trial. Violating any assumption means the geometric distribution does not apply.
How do I find P(X ≥ x) from the table?
Use the complement: P(X ≥ x) = 1 − P(X ≤ x − 1). Look up the CDF value at x − 1 and subtract from 1. For example, P(X ≥ 4) = 1 − P(X ≤ 3). The CDF table directly gives P(X ≤ 3).
Is there a geometric distribution table for all values of p?
Printed tables cover the most common p values (typically p = 0.05 to p = 0.90 in steps). For other p values, use the calculator on this page, which accepts any value between 0 and 1. The formula P(X = x) = (1 − p)x−1 × p works for any p.
How does sample size relate to the geometric distribution?
There is no fixed sample size in the geometric distribution. The number of trials X is the random variable itself — it is not set in advance. This is the key difference from the binomial distribution, where n is fixed before data collection.
What is the geometric distribution CDF formula?
P(X ≤ x) = 1 − (1 − p)x for Convention 1 (X = trial of first success). This gives the probability of obtaining the first success on or before trial x. For Convention 2 (Y = failures before first success): P(Y ≤ y) = 1 − (1 − p)y+1.
When should I use geometric vs Poisson distribution?
Use geometric when you count discrete trials until the first success with a fixed per-trial probability p. Use Poisson when you count events arriving at a continuous rate λ over a time interval, with no fixed trial structure. Poisson does not require a fixed p per event.
What is the standard deviation of the geometric distribution?
The standard deviation is the square root of the variance: SD(X) = √[(1 − p)/p²] = √(q)/p. For p = 0.30, SD(X) = √(0.70/0.09) = √7.78 ≈ 2.79. Higher p values produce smaller standard deviations because the first success tends to arrive sooner and more consistently.
Glossary of Terms
These definitions match the terminology used in the tables and formulas on this page.
| Term | Definition | Formula / Notation | Related Concept |
|---|---|---|---|
| Geometric Distribution | Discrete distribution of trials until first success | X ~ Geom(p) | Negative Binomial (r=1) |
| Bernoulli Trial | Single independent experiment with two outcomes | P(success) = p | Bernoulli Distribution |
| PMF | Probability of exactly x trials | P(X=x) = q^(x-1) × p | CDF |
| CDF | Probability of x or fewer trials | P(X≤x) = 1 − q^x | PMF |
| First Success | The trial that terminates the sequence | X in Convention 1 | Expected Value 1/p |
| Expected Value | Average trial number of first success | E(X) = 1/p | Variance |
| Memoryless Property | Past trials carry no information about future success | P(X>s+t|X>s) = P(X>t) | Exponential Distribution |
| Negative Binomial | Trials until r-th success; geometric when r = 1 | NB(r, p), r = 1 → Geom | Geometric Distribution |
References and External Resources
NIST/SEMATECH e-Handbook of Statistical Methods. Section 1.3.6.8: Geometric Distribution. National Institute of Standards and Technology. itl.nist.gov/div898/handbook — Authoritative U.S. government statistical reference with formula definitions and parameter descriptions.
OpenStax Statistics (2023). Chapter 4: Discrete Random Variables — Geometric Distribution. Rice University. openstax.org — Free, peer-reviewed introductory statistics textbook with worked examples and practice problems.
DeGroot, M. H., & Schervish, M. J. (2012). Probability and Statistics (4th ed.). Pearson. — Graduate-level treatment of the geometric distribution, its relationship to the negative binomial, and the memoryless property proof.
Penn State STAT 414: Probability Theory. Section 3.2: The Geometric and Negative Binomial Distributions. Pennsylvania State University. online.stat.psu.edu/stat414 — Open-access graduate course notes covering both geometric distribution conventions.
Related Statistical Tables & Resources
Understanding the Geometric Distribution: Key Takeaways
Why the PMF Decreases
Every additional trial before the first success requires at least one more failure. Since each failure has probability q < 1, the probability of x − 1 consecutive failures shrinks as x grows. This is why P(X = x) = qx−1 × p decreases monotonically as x increases for any fixed p < 1. The PMF graph always shows a decreasing bar chart, with the tallest bar at x = 1.
How p Affects Table Values
Larger p values produce higher probabilities at small x (early success is more likely) and lower probabilities at large x. When p = 0.90, P(X = 1) = 0.9000 and the distribution concentrates near x = 1. When p = 0.05, P(X = 1) = 0.0500 and the distribution spreads across many values of x. This spread is captured by the variance formula (1 − p)/p², which grows rapidly as p decreases toward 0.
Relationship to the Exponential Distribution
The geometric distribution is the discrete analogue of the continuous exponential distribution. Both share the memoryless property. Both describe waiting times until a first event. The exponential distribution applies when time is continuous and arrivals follow a Poisson process. The geometric distribution applies when time is measured in discrete trial counts. Understanding one aids understanding of the other. See the Poisson distribution page for the connection.