3.2 - Discrete Probability Distributions

This section takes a look at some of the characteristics of discrete random variables.

Consider the data set with the values: \(0, 1, 2, 3, 4\). If \(X\) is a random variable of a random draw from these values, what is the probability you select 2? If we assume the probabilities of each of the values is equal, then the probability would be \(P(X=2)=\frac{1}{5}\). We can define the probabilities of each of the outcomes using the probability mass function (PMF) described in the last section. If we assume the probabilities of all the outcomes were the same, the PMF could be displayed in function form or a table. As a function, it would look like: \(f(x)=\begin{cases} \frac{1}{5} & x=0, 1, 2, 3, 4\\ 0 & \text{otherwise} \end{cases}\)

As a table, it would look like:

\(x\) 0 1 2 3 4
\(f(x)\) 1/5 1/5 1/5 1/5 1/5

Recall that for a PMF, \(f(x)=P(X=x)\). In other words, the PMF gives the probability our random variable is equal to a value, x. We can also find the CDF using the PMF.

Example 3-1: CDF Section

Find the CDF, in tabular form of the random variable, X, as defined above.

\(x\) 0 1 2 3 4
\(f(x) = P(X=x)\) 1/5 1/5 1/5 1/5 1/5


Recall that \(F(X)=P(X\le x)\).  Start by finding the CDF at \(x=0\).

\(F(0)=P(X\le 0)\)

Since 0 is the smallest value of \(X\), then \(F(0)=P(X\le 0)=P(X=0)=\frac{1}{5}\)

Now, find \(F(1)\).

\begin{align} F(1)=P(X\le 1)&=P(X=1)+P(X=0)\\&=\frac{1}{5}+\frac{1}{5}\\&=\frac{2}{5}\end{align}

Next, \(F(2)\).

\begin{align} F(2)=P(X\le 2)&=P(X=2)+P(X=1)+P(X=0)\\&=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\\&=\frac{3}{5}\end{align}

Next, \(F(3)\).

\begin{align} F(3)=P(X\le 3)&=P(X=3)+P(X=2)+P(X=1)+P(X=0)\\&=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\\&=\frac{4}{5}\end{align}

Finally, \(F(4)\).

\begin{align} F(4)=P(X\le 4)&=P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)\\&=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\\&=\frac{5}{5}=1\end{align}


In table form...

\(x\) 0 1 2 3 4
\(F(x) = P(X\le x)\) 1/5 2/5 3/5 4/5 5/5=1

This table provides the probability of each outcome and those prior to it. Thus, the probability for the last event in the cumulative table is 1 since that outcome or any previous outcomes must occur.

Try It! Section

Use the table from the example above to answer the following questions.

\(x\) 0 1 2 3 4
\(f(x) = P(X=x)\) 1/5 1/5 1/5 1/5 1/5
  1. Find \(P(X=1)\)


  2. Find \(P(X\le 2)\)
    \begin{align} P(X\le 2)&=P(X=0)+P(X=1)+P(X=2)\\&=\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\\&=\dfrac{3}{5}\end{align}
  3. \(P(X<3)\)

    \(P(X<3)=P(X\le 2)=\dfrac{3}{5}\).  Note that \(P(X<3)\) does not equal \(P(X\le 3)\) as it does not include \(P(X=3)\).

  4. \(P(1\le X\le 3)\)
    \(P(1\le X\le 3)=P(X=1)+P(X=2)+P(X=3)=\dfrac{3}{5}\)