Geometric Distribution Calculator
Free Geometric distribution Calculator for sequences. Enter values to get step-by-step solutions with formulas and graphs.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
P(X = k) = p * (1 - p)^(k-1)
Where p is the probability of success on each trial, k is the trial number on which the first success occurs, and (1-p) is the probability of failure. The mean is 1/p and the variance is (1-p)/p^2.
Worked Examples
Example 1: Quality Control Inspection
Problem:A factory has a 5% defect rate. What is the probability of finding the first defective item on the 10th inspection?
Solution:P(X = 10) = 0.05 * (1 - 0.05)^(10-1)\nP(X = 10) = 0.05 * 0.95^9\nP(X = 10) = 0.05 * 0.6302\nP(X = 10) = 0.03151\nExpected inspections until first defect: 1/0.05 = 20\nP(finding defect within 10 inspections) = 1 - 0.95^10 = 0.4013
Result:P(X=10) = 3.15% | Expected trials = 20 | P(X<=10) = 40.13%
Example 2: Sales Conversion Rate
Problem:A salesperson converts 20% of leads. What is the probability of making the first sale within 5 calls?
Solution:P(X <= 5) = 1 - (1 - 0.20)^5\nP(X <= 5) = 1 - 0.80^5\nP(X <= 5) = 1 - 0.32768\nP(X <= 5) = 0.67232\nExpected calls until first sale: 1/0.20 = 5\nP(exactly on 5th call) = 0.20 * 0.80^4 = 0.08192
Result:P(within 5 calls) = 67.23% | P(exactly 5th) = 8.19% | Expected = 5 calls
Frequently Asked Questions
What is a geometric distribution and when is it used?
A geometric distribution models the number of independent Bernoulli trials needed to achieve the first success, where each trial has the same probability of success p. It is the discrete analog of the exponential distribution. Common applications include modeling the number of coin flips until the first heads, the number of products inspected until finding a defective one, the number of sales calls before making a sale, and the number of attempts to establish a network connection. The distribution assumes each trial is independent and the probability of success remains constant. It is the only memoryless discrete probability distribution, meaning past failures do not affect future success probability.
What is the probability mass function (PMF) of the geometric distribution?
The PMF gives the probability that the first success occurs on exactly the kth trial: P(X = k) = p * (1 - p)^(k-1), where p is the success probability and k is the trial number (k = 1, 2, 3, ...). This formula makes intuitive sense: you need (k-1) failures each with probability (1-p), followed by one success with probability p. For example, with p = 0.3, the probability of first success on trial 4 is 0.3 * 0.7^3 = 0.3 * 0.343 = 0.1029 or about 10.29%. The PMF decreases geometrically (by a constant ratio of 1-p), which is where the distribution gets its name.
How do you calculate the cumulative distribution function (CDF) of the geometric distribution?
The CDF gives the probability of achieving the first success within k trials: P(X <= k) = 1 - (1 - p)^k. This formula is derived by noting that not succeeding within k trials means failing k times in a row with probability (1-p)^k, so the complement gives the CDF. For example, with p = 0.2, the probability of success within 5 trials is 1 - 0.8^5 = 1 - 0.32768 = 0.67232 or about 67.2%. The CDF is useful for answering questions like how many trials are needed for a given confidence level. To find the number of trials needed for 95% certainty, solve 1 - (1-p)^k >= 0.95 for k.
What are the mean and variance of the geometric distribution?
The mean (expected value) of the geometric distribution is E[X] = 1/p, and the variance is Var(X) = (1-p)/p^2. The standard deviation is sqrt((1-p)/p^2) = sqrt(1-p)/p. For example, if p = 0.25, the expected number of trials until the first success is 1/0.25 = 4, with variance = 0.75/0.0625 = 12 and standard deviation = sqrt(12) = 3.46. The mean being 1/p is intuitive: if you succeed 25% of the time, you expect to need 4 tries on average. As p approaches 1, both mean and variance decrease toward 1 and 0 respectively, since success becomes nearly certain on the first trial.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy