# The Geometric Distribution

Contents

## Summary

• The geometric distribution has a single parameter (p) = X ~ Geo(p)
• Geometric distribution can be written as $P(X\quad =\quad x)\quad =\quad { q }^{ x-1 }p$,
where q = 1 – p
• The mean of the geometric distribution is: $\mu \quad =\quad E(X)\quad =\quad \frac { 1 }{ p }$
• The variance of the geometric distribution is: ${ \sigma }^{ 2 }\quad =\quad var(X)\quad =\quad \frac { 1\quad -\quad p }{ { p }^{ 2 } } \quad =\quad \frac { q }{ { p }^{ 2 } }$
• The standard deviation of the geometric distribution is: $\sigma \quad =\quad .S.D\quad =\quad \sqrt { \frac { 1\quad -\quad p }{ { p }^{ 2 } } } \quad =\quad \sqrt { \frac { q }{ { p }^{ 2 } } }$

The geometric distribution are the trails needed to get the first success in repeated and independent binomial trial. Each trial has two possible outcomes, it can either be a success or a failure. We can write this as:

1. P(Success) = p (probability of success known as p, stays constant from trial to trial).
2. P(failure) = q (probability of failure is a complement of success, thus for any trial it remains 1 – p).

Hence, we can write its probability density function as:

$P(X\quad =\quad x)\quad =\quad { q }^{ x-1 }p$

In other words:

$P(X\quad =\quad x)\quad =\quad { (1\quad -\quad p) }^{ x-1 }p$

For x = 1, 2, 3….

The geometric distribution has a single parameter, the probability of success (p). We can thus write this as:

X ~ Geo(p)

Moreover, the mean and variance are the functions of p.

• The mean of the geometric distribution is:

$\mu \quad =\quad E(X)\quad =\quad \frac { 1 }{ p }$

• The variance of the geometric distribution is:

${ \sigma }^{ 2 }\quad =\quad var(X)\quad =\quad \frac { 1\quad -\quad p }{ { p }^{ 2 } } \quad =\quad \frac { q }{ { p }^{ 2 } }$

• The standard deviation of the geometric distribution is:

$\sigma \quad =\quad .S.D\quad =\quad \sqrt { \frac { 1\quad -\quad p }{ { p }^{ 2 } } } \quad =\quad \sqrt { \frac { q }{ { p }^{ 2 } } }$

#### Example #1

Q. In a large population of school students 30% have received karate training.

If students from this population are randomly selected, calculate:

a) what is the probability that the 6th person that was chosen at randomly was the first student to have received the karate training.

Solution:

Here we require the number of trials needed to get the first success. Since students are randomly selected from a large population, we can say that the trials are independent.

In addition, the fact that the first person has or has not received the karate training, does nothing to solve our query of whether the next randomly selected person has received the training or not.

To sum up what was said above:

1. Independent trails have been repeated
2. We want to know the number of trials required to get the first success
3. The first student who got the karate training
4. And lastly, the probability of success is constant

We can hence, say:

P = 0.3 (30%)

We want the probability of randomly selected 6 students thus we will write:

$P(X\quad =\quad 6)\quad =\quad { (1\quad -\quad 0.3) }^{ 5 }(0.3)$

Note: We want the first 5 students to not have received the karate training and the 6 to have.

Thus, calculating the above equation we get:

$P(X\quad =\quad 6)\quad =\quad { (1\quad -\quad 0.3) }^{ 5 }(0.3)$

$\approx \quad 0.0504$  (when rounded off)

We can thus conclude the answer by saying;

The probability that the 6th person that was chosen randomly was the first student to have received the karate training is 0.0504.

b) Find the mean, variance and the standard deviation of the example above.

Solution:

We will use the formulas required to find the three of them.

For mean:

$\mu \quad =\quad \frac { 1 }{ p } \quad =\quad \frac { 1 }{ 0.3 } \quad =\quad 3.3$

For the variance:

${ \sigma }^{ 2 }\quad =\quad \frac { 1\quad -\quad 0.3 }{ { 0.3 }^{ 2 } } \quad =\quad 7.7$

Lastly, for the standard deviation:

$\sigma \quad =\quad \sqrt { 7.7 } \quad =\quad 2.79$

c) What is the probability that the first student trained in karate occurs on or before the 3rd person was sampled.

Solution:

Now, we need to find the probability that the random variable X is less than equal to 3.

We will write this as:

$P(X\quad \le \quad 3)\quad =\quad P(X\quad =\quad 1)\quad +\quad P(X\quad =\quad 2)\quad +\quad P(X\quad =\quad 3)$

Using, this formula, we can say:

${ (1\quad -\quad p) }^{ x-1 }p$

$P(X\quad \le \quad 3)\quad =\quad { (0.7) }^{ 1-0 }(0.3)\quad +\quad { (0.7) }^{ 2-1 }(0.3)\quad +\quad { (0.7) }^{ 3-1 }(0.3)$

= 0.657

In a nutshell, we can say that the probability that the first student trained in karate occurs on or before the 3rd person was sampled is 0.657.