5.4.3 The Geometric Distribution

In this topic we will learn how to:

use the formula for probabilities for the geometric distribution, and recognise practical situations where this distribution is a suitable model
use formulae for the expectation and variance of the geometric distribution

The geometric distribution is a discrete distribution. This means it takes exact values only. It represents the number of trials up to and until the first success. For a random variable to follow a geometric distribution, the following conditions have to be satisfied:

There are only two possible outcomes. Success and Failure.
There are a series of trials. Number of trials is not fixed.
Trials are independent of each other
Trials are independent of each other
The probability of success is constant

Note: The only difference between conditions for a binomial distribution and those of a geometric distribution is that binomial distribution has a fixed number of trials and geometric has a series of trials.

If the following four conditions are met, then the random variable follows a geometric distribution.

The notation for a geometric distribution is as follows,
$X \sim Geo(p)$ Which reads ‘ $X$ follows a geometric distribution, with the parameter $p$ ‘. $X$ represents a random variable, $Geo$ represents a geometric distribution, $p$ represents the probability of success.
The formula for geometric probabilities is,
$P(X = r) = p(1 - p)^{r - 1}$ The above may also be written as,
$P_{r} = p(1 - p)^{r - 1}$ In other literature, $1 - p$ , may be written as $q$ .

There are some shortcuts for calculating some geometrical probabilities, to make it less tedious,
$P(X \le r) = 1 - q^{r}$ $P(X > r) = q^{r}$ Note: Remember that $q$ is $1 - p$ .

Under geometric conditions we can also calculate the mean. The formula for mean is,
$E(X) = \frac{1}{p}$ Let’s look at some past paper questions.

1. A fair spinner with $5$ sides numbered $1$ , $2$ , $3$ , $4$ , $5$ is spun repeatedly. The score on each spin is the number on the side on which the spinner lands. (9709/52/F/M/21 number 1)

(a) Find the probability that a score of $3$ is obtained for the first time on the $8$ th spin.

We can tell that this is a geometric distribution because we’re spinning the spinner up until the first success. So let’s define the random variable,
$X \textmd{- r.v, number of spins up until a }3\textmd{ is first obtained}$ State the distribution and parameters of the random variable,
$X \sim Geo\left(\frac{1}{5}\right)$ Note: Since the spinner is fair, there is a $1$ in $5$ chance of getting a $3$ .

The question requires us to find the probability that the first success is at the $8$ trial,
$P(X = 8)$ Let’s use the formula for geometric probabilities,
$P(X = r) = p(1 - p)^{r - 1}$ Substitute into the formula,
$P(X = 8) = \left(\frac{1}{5}\right)\left(1 - \frac{1}{5}\right)^{8 - 1}$ This simplifies to give,
$P(X = 8) = 0.0419$ Therefore, the final answer is,
$P(X = 8) = 0.0419$ (b) Find the probability that fewer than $6$ spins are required to obtain a score of $3$ for the first time.

Let’s write out, mathematically, what the question is asking us to do,
$P(X < 6)$ This can be written as,
$P(X < 6) = P(X = 5, 4, 3, 2, 1)$ Note: We do not include $0$ under a geometric distribution, because the distribution is the number of trials up until a first success, so there has to be at least one trial.

We could go ahead and evaluate all the probabilities above but that would be tedious. Instead, we could use one of the shortcuts outlined earlier,
$P(X \le r) = 1 - q^{r}$ We can rewrite our problem as,
$P(X < 6) = P(X \le 5)$ Note: Since this is a discrete distribution, less than $6$ means less than or equal to $5$ .

Now we can use the shortcut,
$P(X \le r) = 1 - q^{r}$ $P(X \le 5) = 1 - \left(1 - \frac{1}{5}\right)^{5}$ This simplifies to give,
$P(X \le 5) = \frac{2101}{3125}$ $P(X \le 5) = 0.672$ Therefore, the final answer is,
$P(X \le 5) = 0.672$ 2. Two fair coins are thrown at the same time. The random variable $X$ is the number of throws of the two coins required to obtain two tails at the same time. (9709/51/O/N/21 number 1)

(a) Find the probability that the two tails are obtained for the first time on the $7$ th throw.

We can tell that this is a geometric distribution because we’re tossing the coins up until the first success. The random variable has already been defined for us, so let’s state its distribution and parameters,
$X \sim Geo(p)$ $p$ is the probability of obtaining two tails at the same time when two fair coins are thrown at the same time,
$p = (0.5)^{2}$ $p = 0.25$ $X \sim Geo(0.25)$ The question requires us to find the probability that the first success is on the $7$ th throw,
$P(X = 7)$ Let’s use the formula for geometric probabilities,
$P(X = r) = p(1 - p)^{r - 1}$ $P(X = 7) = (0.25)(1 - 0.25)^{7 - 1}$ This simplifies to give,
$P(X = 7) = \frac{729}{16\ 384}$ $P(X = 7) = 0.0445$ Therefore, the final answer is,
$P(X = 7) = 0.0445$ (b) Find the probability that it takes more than $9$ throws to obtain two tails for the first time.

The question requires us to find the probability that it takes more than $9$ throws to get the first success,
$P(X > 9)$ We can use one of the shortcuts to solve this,
$P(X > r) = q^{r}$ $P(X > 9) = (1 - 0.25)^{9}$ Note: Remember that $q = 1- p$ .

This simplifies to give,
$P(X > 9) = 0.0751$ Therefore, the final answer is,
$P(X > 9) = 0.0751$ 3. An ordinary fair die is thrown repeatedly until a $5$ is obtained. The number of throws taken is denoted by the random variable $X$ . (9709/52/M/J/21 number 1)

(a) Write down the mean of $X$ .

The random variable $X$ has already been defined for us. Let’s state the distribution and parameters of $X$ . Since they are throwing the die up until the first success, it is a geometric distribution,
$X \sim Geo\left(\frac{1}{6}\right)$ Note: It’s a fair die, so the probability of rolling a $5$ is $1$ in $6$ , hence $p = \frac{1}{6}$ .

To find the mean, under a geometric distribution, we use the formula,
$E(X) = \frac{1}{p}$ Substitute $p$ in the formula,
$E(X) = \frac{1}{\frac{1}{6}}$ This simplifies to give,
$E(X) = 6$ Therefore, the final answer is,
$E(X) = 6$ (b) Find the probability that a $5$ is first obtained after the $3$ rd throw but before the $8$ th throw.

Let’s write out the question mathematically,
$P(3 < X < 8)$ This can be written as,
$P(3 < X < 8) = P(X = 4, 5, 6, 7)$ Note: Alternatively, you could split this into two inequalities and use the shortcuts.

Let’s use the formula for geometric probabilities to evaluate,
$P(X = r) = p(1 - p)^{r - 1}$ $P(X = 4) = \left(\frac{1}{6}\right)\left(1 - \frac{1}{6}\right)^{4 - 1}$ $P(X = 4) = \frac{125}{1\ 296}$ $P(X = 5) = \left(\frac{1}{6}\right)\left(1 - \frac{1}{6}\right)^{5 - 1}$ $P(X = 5) = \frac{625}{7\ 776}$ $P(X = 6) = \left(\frac{1}{6}\right)\left(1 - \frac{1}{6}\right)^{6 - 1}$ $P(X = 6) = \frac{3\ 125}{46\ 656}$ $P(X = 7) = \left(\frac{1}{6}\right)\left(1 - \frac{1}{6}\right)^{7 - 1}$ $P(X = 7) = 0.05581632945$ Now let’s go back to our equation,
$P(3 < X < 8) = P(X = 4, 5, 6, 7)$ $P(3 < X < 8) = \frac{125}{1\ 296} + \frac{625}{7\ 776} + \frac{3\ 125}{46\ 656} + 0.05581632945$ This simplifies to give,
$P(3 < X < 8) = 0.300$ Therefore, the final answer is,
$P(3 < X < 8) = 0.300$ (c) Find the probability that a $5$ is first obtained in fewer than $10$ throws.

Let’s write out the problem mathematically,
$P(X < 10)$ We can rewrite this as,
$P(X < 10) = P(X \le 9)$ We can use a shortcut to solve that,
$P(X \le r) = 1 - q^{r}$ $P(X \le 9) = 1 - \left(\frac{5}{6}\right)^{9}$ $P(X \le 9) = 0.806$ Therefore, the final answer is,
$P(X \le 9) = 0.806$