The geometric distribution is either of two discrete probability distributions:

- The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, …}
- The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, … }

$\large P(X=x)=p(1-p)^{x-1};x=1,2,3,….$

$\large P(x)=o; other\;wise$

$\large P(X=x)=p(1-p)^{x};x=0,1,2,3,….$

$\large P(x)=o; other\;wise$

Where,

$p$ is the probability of occurrence

Mean and variance can be found using the value of p.

Mean = $\frac{1}{p}$

Variance = $\frac{1-p}{p^{2}}$

### Solved Example

**Question: **Calculate the probability density of geometric distribution if the value of p is 0.42; x = 1,2,3,…….., also find out the mean and variance.

**Solution:**

Given that, p = 0.42 and the value of x is 1,2,3,……………

Formula for the probability density of geometric distribution function,

P(x) = p$(1-p)^{x-1}$; x = 1,2,3,…

P(x) = 0; other wise

P(x) = 0.42$(1-0.42)^{x-1}$

P(x) = 0 other wise

Mean = $\frac{1}{p}$= $\frac{1}{0.42}$ = 2.380

Variance = $\frac{1-p}{p^{2}}$

Variance = $\frac{1-0.42}{0.42^{2}}$

Variance = 3.288