Introduction to Discrete Probability Distributions (Business)

What is a Discrete Probability distribution?

The probability distribution of a discrete random variable $X$ is the set $S$ of all the possible outcomes $X$ can take and corresponding probabilities $\mathrm{P}(x)$ (or equivalently $\mathrm{P}(X=x)$) for each $x \in S$ (each value of $x$ in the set $S$). Remember, for a discrete probability distribution to be valid, the probabilities must sum to one.

Illustrative Example 1

Suppose that exactly 4 loan applications are reviewed each day by a small bank branch. Suppose further that these loans are accepted independently of each other so the probability that one of the four loans reviewed in a given day is accepted does not affect the probability that one of the other four loans is accepted. We have the following probabilities:

• No applications are accepted with probability $0.2$,
• $1$ application is accepted with probability $0.25$
• $2$ applications are accepted with probability $0.3$,
• $3$ applications are accepted with probability $0.15$ and
• $4$ are accepted with probability $0.1$.

This is an example of a discrete probability distribution and is valid because the probabilities sum to 1 ($0.2+0.25+0.3+0.15+0.1=1$). We can present this probability distribution as a table:

Cumulative Probabilities

If $X$ is a random variable then the cumulative probability ($\mathrm{P}(X \leq 4)$ for example) is the probability that $X$ takes any value less than or equal to 4.

If $X$ is a discrete random variable taking only positive values, then we can obtain this cumulative probability by summing each of the probabilities $\mathrm{P}(X=j)$ for $j=0, 1, 2, 3.$. That is,

\[\mathrm{P}(X \leq 4)=\mathrm{P}(X=0)+\mathrm{P}(X=1)+\mathrm{P}(X=2)+\mathrm{P}(X=3).\] Note: Since the sum of all of the probabilities must be 1, we have for any value $j$:

\[\mathrm{P}(X \leq j)=1-\mathrm{P}(X \gt j).\]

Thus if $\mathrm{P}(X \leq j)$ is more straightforward to calculate than $\mathrm{P}(X \gt j)$ (or vice versa), we can use the above fact to obtain one from the other. We will see examples of this below.

The Expectation and Variance

The expectation $\mathrm{E}[X]$ of a discrete random variable $X$ is the sum of each of the possible outcomes multiplied by their associated probabilities.

The variance $\mathrm{Var}(X)$ of a discrete random variable $X$ is defined as:

However it can be simplified to this expression:

where $\mathrm{E}[X^2]$ is the sum of each outcome squared and multiplied by its corresponding probability:

\[\sum\limits_{x \in S}{x^2 \times \mathrm{P}(x)}\]

and $\mathrm{E}[X]^2$ is the mean, $\mathrm{E}[X]$, squared:

\[\left(\sum\limits_{x \in S}{x \times \mathrm{P}(x)}\right)^2\].

Note: We also denote $\mathrm{E}[X]$ as $\mu$.

Worked Example 1

Combinations

A combination is a way of selecting a number of items from a collection of items where the order of selection is not important. Combinations can be used to calculate probabilities, which we shall demonstrate with an example.

Illustrative Example 2

Imagine you are about to be shipwrecked on a deserted island. There is some rope, a knife, a box of matches and a blanket, but you can only take two items with you before the ship sinks. How many different ways are there of choosing two items? The different ways of choosing two items are combinations (since order is not important) and are listed below:

1. Rope & knife
2. Rope & matches
3. Rope & blanket
4. Knife & matches
5. Knife & blanket
6. Matches & blanket.

So, there are six different ways of selecting two items.

Now imagine the items are all in unmarked identical wooden boxes, so you do not know which items you are selecting. What is the probability you have selected the knife and matches?

\begin{equation} \mathrm{P}(\text{select knife and matches}) = \dfrac{\text{Number of ways of selecting the knife and matches}}{\text{Total number of combinations}} =\dfrac{1}{6} \end{equation}

With this number of items, it is quite easy to list all the possible selections. However, for larger numbers of items listing all of the possible selections can become tedious and time consuming. Fortunately there is a much faster way of calculating such combinations. The formula is:

where $n!$ denotes the factorial of $n$ ($n!=n\times(n-1)\times(n-2)\times...\times1$), $n$ is the total number of items (e.g. people. numbers, objects...), $r$ is how many items we want to “choose” and $r \leq n$. We pronounce this formula as “$n$ C $r$” or “$n$ choose $r$”.

We can now use this formula to calculate how many ways there are of selecting six items from twelve as follows: \begin{align} {}^{12}\mathrm{C}_6 &=\dfrac{12!}{6!(12-6)!}\\ &=\dfrac{12\times11\times\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{(6\times5\times4\times3\times2\times1)\times(6\times5\times4\times3\times2\times1)}\\ &=\dfrac{479001600}{720\times720}\\ &=924. \end{align}

Selecting lottery numbers is an example of choosing a combination. A person must pick $6$ numbers from $49$. With each draw there is only one winning combination. But how many combinations are possible?

There are ${}^{49}\mathrm{C}_6 = 13983816$ possible combinations or ways of picking $6$ numbers from the $49$ possible numbers where order doesn't matter. This means that the probability of winning the lottery is $\dfrac{1}{13983816}.$

Permutations

A permutation is a way of selecting a specified number of items from a collection of items where the order of selection is important. Another way of thinking about this is that it is a way of arranging a number of items.

We use the formula:

where $n!$ denotes the factorial of $n$ ($n!=n\times(n-1)\times(n-2)\times...\times1$), $n$ is the total number of items (e.g. people. numbers, objects...), $r$ is how many items we want to arrange from the $n$ items and $r \leq n$. where $n$ is the total number of things, $r$ is how many things we are choosing and $r \leq n$. As with combinations, we can also use permutations to help calculate probabilities.

Worked Example

Test Yourself

To practice questions on the Binomial distribution and Poisson distribution click the following link:

Numbas test.

See Also

For information on continuous probability distributions see the pages on continuous probability distributions.