Geometric Distribution Examples
- 1 Geometric Distribution Calculator
- 2 How to find Geometric Distribution Probabilities?
- 3 Definition of Geometric Distribution
- 4 Distribution Function of Geometric Distribution
- 5 Mean of Geometric Distribution
- 6 Variance of Geometric Distribution
- 7 Geometric Distribution Example 1
- 8 Geometric Distribution Example 2
- 9 Conclusion
Geometric Distribution Calculator
Geometric distribution calculator is used to find the probability and cumulative probabilities for geometric random variable given the probability of success ($p$).
Geometric Distribution Calculator | |
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Probability of success (p): | |
No. of failure before first success (x): | |
Result | |
Probability : P(X = x) | |
Cumulative Probability : P(X ≤ x) | |
Cumulative Probability : P(X < x) | |
Cumulative Probability : P(X ≥ x) | |
Cumulative Probability : P(X > x) | |
How to find Geometric Distribution Probabilities?
Step 1 – Enter the probability of success $p$
Step 2 – Enter the value of no. of failure before first success $x$
Step 3 – Click on "Calculate" button to get geometric distribution probabilities
Step 4 – Gives the output probability at $x$ for geometric distribution
Step 5 – Gives the output cumulative probabilities for geometric distribution
Definition of Geometric Distribution
A discrete random variable $X$ is said to have geometric
distribution with parameter $p$ if its probability mass function is given by
$$ \begin{align*} P(X=x)&= \begin{cases} q^x p, & x=0,1,2,\ldots \\ & 0 < p < 1, q=1- p \\ 0, & Otherwise. \end{cases} \end{align*} $$
Distribution Function of Geometric Distribution
The distribution function of geometric distribution is $F(x)=1-q^{x+1}, x=0,1,2,\cdots$.
Mean of Geometric Distribution
The mean of Geometric distribution is $E(X)=\dfrac{q}{p}$.
Variance of Geometric Distribution
The variance of Geometric distribution is $V(X)=\dfrac{q}{p^2}$.
Geometric Distribution Example 1
The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8. Assume the trials are independent.
Compute the probability that the first successful alignment
a. requires exactly four trials,
b. requires at most three trials,
c. requires at least three trials.
Solution
Let $X$ denote the number of trials required for first successful optical alignment. Thus the random variable $X$ take values $X=1,2,3,\cdots$.
Given that the probability of succcessful optical alignment in the assembly of an optical data storage product is $p=0.8$. The trials are independent.
Thus random variable $X$ follows a geometric distribution with probability mass function
$$ \begin{aligned} P(X=x)&= p(1-p)^{x-1}; \; x=1,2,\cdots\\ &= 0.8 (1-0.8)^{x-1}\; x=1,2,\cdots\\ &= 0.8 (0.2)^{x-1}\; x=1,2,\cdots \end{aligned} $$
a. The probability that the first successful alignment requires exactly $4$ trials is
$$ \begin{aligned} P(X=4)&= 0.8(0.2)^{4 -1}\\ &= 0.8(0.008)\\ &= 0.0064. \end{aligned} $$
b. The probability that the first successful alignment requires at most $3$ trials is
$$ \begin{aligned} P(X\leq 3)&= \sum_{x=1}^{3}P(X=x)\\ &= P(X=1)+P(X=2)+P(X=3)\\ &= 0.8+0.16+0.032\\ &= 0.992. \end{aligned} $$
c. The probability that the first successful alignment requires at least $3$ trials is
$$ \begin{aligned} P(X\geq 3)&= 1-P(X\leq 2)\\ &= 1- \sum_{x=1}^{2}P(X=x)\\ &= 1-\big(P(X=1)+P(X=2)\big)\\ &= 1-\big(0.8+0.16\big)\\ &= 1-0.96\\ &= 0.04. \end{aligned} $$
Geometric Distribution Example 2
An old gas water heater has a pilot light which much be lit manually, using a match. The probability of successfully lighting the pilot light on any given attempt is 82%.
a. Compute the probability that it takes no more than 4 tries to light the pilot light.
b. Compute the probability that the pilot light is lit on the 5th try.
c. Compute the probability that it takes more than four tries to light the pliot light.
Solution
Let $X$ denote the number of attempts to light (success) the pilot light.
Given that $p=0.82$ is the probability of successfully lighting the pilot light on any given attempt.
Thus random variable $X$ follows a geometric distribution with probability mass function
$$ \begin{aligned} P(X=x)&= p(1-p)^{x-1}; \; x=1,2,\cdots\\ &= 0.82 (1-0.82)^{x-1}\; x=1,2,\cdots\\ &= 0.82 (0.18)^{x-1}\; x=1,2,\cdots \end{aligned} $$
a. The probability that it takes no more than 4 tries to light the pilot light.
$$ \begin{aligned} P(X\leq 4)&= F(4)\\ &=1-q^{4}\\ &=1- 0.18^{4}\\ &=1-0.001\\ &=0.999 \end{aligned} $$
b. The probability that the pilot light is lit on the 5th try
$$ \begin{aligned} P(X=5)&= 0.82(0.18)^{5 -1}\\ &= 0.82(0.001)\\ &= 0.0009. \end{aligned} $$
c. The probability that it takes more than four tries to light the pliot light
$$ \begin{aligned} P(X> 4)&= 1-P(X\leq 4)\\ &= 1- 0.999\\ &= 0.001 \end{aligned} $$
Conclusion
In this tutorial, you learned about how to calculate mean, variance and probabilities of geometric distribution. You also learned about how to solve numerical problems based on geometric distribution.
To read more about the step by step tutorial on Geometric distribution refer the link Geometric Distribution. This tutorial will help you to understand Geometric distribution and you will learn how to derive mean of geometric distribution, variance of geometric distribution, moment generating function and other properties of geometric distribution.
To learn more about other discrete probability distributions, please refer to the following tutorial:
Let me know in the comments if you have any questions on Geometric Distribution Examples and your thought on this article.