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, … }

Where,

is the probability of occurrence
Mean and variance can be found using the value of p.
Mean =

$\begin{array}{l}\frac{1}{p}\end{array}$

Variance =

$\begin{array}{l}\frac{1-p}{p^2}\end{array}$

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

; x = 1,2,3,…
P(x) = 0; other wise

P(x) = 0.42

P(x) = 0 other wise

Mean =

=

$\begin{array}{l}\frac{1}{0.42}\end{array}$

= 2.380

Variance =

Variance =

Variance = 3.288