Geometric Distribution Formula
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, … }
\(\begin{array}{l}\large P(X=x)=p(1-p)^{x-1};x=1,2,3,….\end{array} \)
\(\begin{array}{l}\large P(x)=o; other\;wise\end{array} \)
\(\begin{array}{l}\large P(X=x)=p(1-p)^{x};x=0,1,2,3,….\end{array} \)
\(\begin{array}{l}\large P(x)=o; other\;wise\end{array} \)
Where,
\(\begin{array}{l}p\end{array} \)
is the probability of occurrenceMean 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
\(\begin{array}{l}(1-p)^{x-1}\end{array} \)
; x = 1,2,3,…P(x) = 0; other wise
P(x) = 0.42
\(\begin{array}{l}(1-0.42)^{x-1}\end{array} \)
P(x) = 0 other wise
Mean = \(\begin{array}{l}\frac{1}{p}\end{array} \)= \(\begin{array}{l}\frac{1}{0.42}\end{array} \) = 2.380
Variance = \(\begin{array}{l}\frac{1-p}{p^{2}}\end{array} \)
Variance = \(\begin{array}{l}\frac{1-0.42}{0.42^{2}}\end{array} \)
Variance = 3.288
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