Generated on: June 13, 2026 at 06:01 PM

Geometric Distribution Calculator

Inputs

Success probability	0.3
Trial number	3

Visualization

PMF value 0

Geometric Distribution Calculator

Calculate the probability that the first success occurs on trial k, plus the cumulative probability, mean, and variance of a geometric distribution.

Parameters

Success probability

0 – 1

Trial number

≥ 1

Enter a value to see results.

Exact probability

Details

Cumulative probability

Standard deviation

Expected number of trials

Variance

Probabilities

P(X = k) — probability of first success on trial k

\begin{aligned} P(X=k) &= (1-p)^{k-1} \cdot p \\ &= (1-0.3)^{ 3-1 } \cdot 0.3 \\ &= ? \end{aligned}

P(X ≤ k) — cumulative probability

\begin{aligned} P(X \leq k) &= 1-(1-p)^{k} \\ &= 1-(1-0.3)^{ 3 } \\ &= ? \end{aligned}

Mean (expected number of trials)

\begin{aligned} \mu &= \dfrac{1}{p} \\ &= \dfrac{1}{ 0.3 } \\ &= ? \end{aligned}

Variance

\begin{aligned} \sigma^2 &= \dfrac{1-p}{p^2} \\ &= \dfrac{1-0.3}{ 0.3^2 } \\ &= ? \end{aligned}

Distribution

PMF value 0

Last updated: 2026-06-04

The geometric distribution is a discrete probability distribution that models the number of independent Bernoulli trials required to obtain the first success. Each trial shares the same success probability p and is independent of all others. This distribution arises in reliability analysis, quality control, sales modelling, and any process where repeated attempts are made until a target outcome is reached.

Formula

This calculator uses the trials-until-first-success convention: X is the trial number of the first success, so X ≥ 1. The probability mass function (PMF) and cumulative distribution function (CDF) are:

P(X = k) = (1-p)^{k-1} \cdot p, \quad k = 1, 2, 3, \ldots

P(X \leq k) = 1 - (1-p)^k

The mean and variance of X under this convention are:

\mu = \frac{1}{p}, \qquad \sigma^2 = \frac{1-p}{p^2}

A second convention, sometimes found in combinatorics texts, defines X as the number of failures before the first success (X ≥ 0). The PMF becomes P(X = k) = (1−p)^k · p, the mean is (1−p)/p, and the variance is the same. This calculator does not use that convention.

Worked example

A sales representative closes a deal on roughly 30 % of all client calls, independently call by call. What is the probability that the first deal is closed on exactly the third call?

Using p = 0.30 and k = 3:

P(X = 3) = (1 - 0.30)^{3-1} \times 0.30 = 0.70^{2} \times 0.30 = 0.49 \times 0.30 = 0.147

There is a 14.7 % chance that the third call produces the first deal. The cumulative probability of closing within three calls is:

P(X \leq 3) = 1 - 0.70^3 = 1 - 0.343 = 0.657

So there is a 65.7 % chance that the first success occurs on or before the third attempt. The expected number of calls needed is μ = 1/0.30 ≈ 3.33.

Formula derivation

For the first success to occur on trial k exactly:

Trials 1 through k−1 must all fail. Each fails with probability 1−p, and the trials are independent, so the probability of k−1 consecutive failures is (1−p)^(k−1).
Trial k must succeed, which occurs with probability p.

Multiplying these independent probabilities gives (1−p)^(k−1) · p.

The CDF follows from the complement rule. The only way P(X ≤ k) fails is if all k trials fail, which has probability (1−p)^k. Therefore P(X ≤ k) = 1 − (1−p)^k.

Memoryless property

The geometric distribution is the only discrete distribution with the memoryless property: given that the first success has not yet occurred after m trials, the remaining waiting time has exactly the same geometric distribution as the original. Formally, P(X > m + n | X > m) = P(X > n). This makes the geometric distribution analytically tractable in queuing theory and sequential testing.

Relationship to other distributions

The geometric distribution is a special case of the negative binomial distribution with the number of required successes set to r = 1. More broadly, the sum of r independent geometric random variables follows a negative binomial distribution, which gives the number of trials needed to achieve r successes.

If the success probability p is small and the number of trials n is large, the number of successes in n geometric trials is approximately Poisson-distributed. For large p near 1, the distribution concentrates near k = 1 and becomes increasingly skewed.

Practical applications

Call centres and sales: estimating the average number of contacts required before a conversion.
Reliability and quality control: modelling the number of units inspected before the first defective item is found.
Network retransmission: analysing the number of transmission attempts needed before a packet is received without error.
Clinical trials: computing the expected number of patient screenings until the first eligible enrolment under a fixed inclusion probability.

Frequently Asked Questions (FAQ)

What is the geometric distribution?

The geometric distribution models the number of independent Bernoulli trials needed to obtain the first success. Each trial has the same probability p of success and is independent of all others. For example, if p = 0.3, the distribution describes how many die rolls (with probability 0.3 of landing on the target face) are required until the first success.

The distribution is memoryless: the number of additional trials needed to succeed is the same regardless of how many failures have already occurred.

Which convention does this calculator use?

This calculator uses the trials-until-first-success convention: X is the trial number of the first success, so X ≥ 1. Under this convention, P(X = k) = (1−p)^(k−1) · p. An alternative convention defines X as the number of failures before the first success (X ≥ 0), giving P(X = k) = (1−p)^k · p. The two conventions differ by one unit in the index. This calculator's mean is 1/p; the failures-before-success convention yields a mean of (1−p)/p.

How many trials are needed on average?

The expected number of trials is μ = 1/p. For a success probability of p = 0.25, the mean is 4 trials. For p = 0.1, the mean is 10 trials. The standard deviation is σ = √((1−p)/p²), so at p = 0.25 the standard deviation is √12 ≈ 3.46, indicating substantial trial-to-trial variability. The distribution is right-skewed: most runs end quickly, but rare long runs pull the mean upward.

How does the geometric distribution relate to the negative binomial?

The geometric distribution is a special case of the negative binomial distribution with the number of required successes set to r = 1. The negative binomial with r = 1 gives P(X = k) = (1−p)^(k−1) · p, which is identical to the geometric PMF. The negative binomial generalises this to the number of trials needed to achieve r successes. This relationship is useful when extending geometric models to scenarios where multiple successes are required.

Recommended Next

Binomial Probability Calculator

Calculate P(X=k), P(X≤k), and P(X≥k) for a binomial distribution. Enter the number of trials, successes, and probability per trial.

Combination Calculator — C(n, r)

Calculate combinations C(n, r): the number of ways to choose r items from n when order does not matter. Covers n up to 20.

Descriptive Statistics Calculator

Calculate mean, standard deviation, variance, range, min, and max for 8 data values. Shows both population and sample statistics.

200+ calculators · 10 languages · 100% free

Was this calculator helpful?

Powered by OneCalc ↗