## Geometric Distribution

In this tutorial, we will provide you step by step solution to some numerical examples on geometric distribution to make sure you understand the geometric distribution clearly and correctly.

## 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
```
$$
\begin{equation*}
P(X=x) =\left\{
\begin{array}{ll}
q^x p, & \hbox{$x=0,1,2,\ldots$} \\
& \hbox{$0<p<1$, $q=1-p$} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

## 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 $V(X)=\dfrac{q}{p^2}$.

## 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
```
$$
\begin{aligned}
P(X\geq 3)&= 1-P(X\leq 2)\\
&= 1- \sum_{x=1}^{2}P(X=x)\\
&= 1-\big(P(X=1)+P(X=2)\big)\\
&= 1-\big(0.8+0.16\big)\\
&= 1-0.96\\
&= 0.04.
\end{aligned}
$$
```

## 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

```
$$
\begin{aligned}
P(X> 4)&= 1-P(X\leq 4)\\
&= 1- 0.999\\
&= 0.001
\end{aligned}
$$
```

Hope this tutorial helps you understand how to solve the probalmes based on Geometric distribution.