Geometric distribution calculator

The formulas

X = number of trials up to and including the first success (k = 1, 2, 3, …):

P(X = k) = (1 − p)^k−1 p P(X ≤ k) = 1 − (1 − p)^k mean = 1/p variance = (1 − p)/p²

Y = number of failures before the first success (k = 0, 1, 2, …):

P(Y = k) = (1 − p)^k p P(Y ≤ k) = 1 − (1 − p)^k+1 mean = (1 − p)/p variance = (1 − p)/p²

Worked example

A salesperson closes 20% of calls, independently. What is the probability that the first sale comes on the third call?

P(X = 3) = 0.8² × 0.2 = 0.64 × 0.2 = 0.128

Two failures then a success: 12.8%. The chance of a sale within the first three calls is 1 − 0.8³ = 0.488, and the chance of needing three or more calls is P(X ≥ 3) = 0.8² = 0.64. On average the first sale takes 1/0.2 = 5 calls, with a standard deviation of √20 = 4.47214. The median is 4 calls, since P(X ≤ 3) = 0.488 falls just short of one half and P(X ≤ 4) = 0.5904 passes it. The mode is always 1: a success on the very first trial is the single most likely outcome.

In the failures convention the same event is “2 failures before the first sale”, so P(Y = 2) = 0.8² × 0.2 = 0.128, the same number with k shifted down by one. The mean becomes 0.8/0.2 = 4 failures.

Reading the result

The bars fall by the same factor, 1 − p, from each k to the next, so the chart always decays geometrically, which is where the name comes from. Solid bars mark the values included in the probability you asked for. Open the probability table for P(X = k), P(X ≤ k) and P(X ≥ k) at every value in the bulk of the distribution.

Common mistakes

• Mixing conventions. An answer that is off by one trial almost always comes from this.
• Using the binomial formula. The binomial fixes the number of trials; the geometric fixes the number of successes (one) and lets the trials vary.
• Adding the terms one by one for P(X ≤ k). The closed form 1 − (1 − p)^k is exact and quicker.
• Forgetting independence. If each failure changes the next chance (learning, fatigue), the geometric does not apply.