Geométrica P(X=k)

Geometric P(X = k): trials until first success

The geometric distribution tells you how many independent Bernoulli trials it takes to land the first success, when each trial succeeds with probability p: P(X = k) = (1 − p)^(k − 1) · p, for k ∈ {1, 2, 3, …}. Its mean is 1/p and its variance is (1 − p)/p². Among the discrete distributions, it is the only one that is memoryless: how many times you have already failed says nothing about what comes next. Say you flip a fair coin until heads shows up. The chance it first happens on flip 3 is 0.5·0.5·0.5 = 0.125.

Applications

Reliability engineers use it for the waiting time until the first failure. In quality control it gives the position of the first defective item in a sample. It also covers everyday cases like the number of rolls until a die shows a 6, the attempts before a network connection goes through, or how many job applications you send before the first yes.

FAQ

Does k start at 0 or 1? You will see both conventions in textbooks. Here X is the trial number on which the first success lands, so k ≥ 1. The other version counts the failures that came before the success and starts at 0.

What does memoryless mean? P(X > m + n | X > m) = P(X > n). Once m failures are behind you, the trials still ahead follow the very same distribution as if you were starting over. Luck does not build up.

How does it relate to the negative binomial? The geometric is just the negative binomial with r = 1, the case where you wait for a single success. Add up r independent geometrics and you get a negative binomial.

What if p = 0? Then the first success never comes and the distribution is degenerate. At the other extreme, p = 1 means the first trial always succeeds, so P(X = 1) = 1.