The Big Idea
A discrete probability distribution function (or PDF) is a function which assigns a probability $P(X=x)$ to every value $x$ which $X$ can be equal to.
The "discrete" part means that $X$ can take on a discrete set of values. That is, all possible values of $X$ are either finite, or do not occur over a continuous interval of real numbers.
Example: Roll a fair die. Let $X$ be the possible value of each face. Then the PDF is $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{6}\\ \hline 2 & \frac{1}{6} \\ \hline 3 & \frac{1}{6} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{6} \\ \hline 6 & \frac{1}{6} \\ \hline \end{array} $$
Long-Term Probabilities of Rolling a Fair Die
| Empirical Probability | |||
| Roll Result |
Huge Fact: A PDF table is nothing more than a loooooooooong term frequency table.
For example, if you had eternity to roll a fair die and record the results in a frequency table, it would be $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{6}=0.1\overline{6}\\ \hline 2 & \frac{1}{6}=0.1\overline{6} \\ \hline 3 & \frac{1}{6}=0.1\overline{6} \\ \hline 4 & \frac{1}{6}=0.1\overline{6} \\ \hline 5 & \frac{1}{6}=0.1\overline{6} \\ \hline 6 & \frac{1}{6}=0.1\overline{6} \\ \hline \end{array} $$ All the stuff that you can do with frequency tables you can do with PDF tables!
Example: A crooked die has the following PDF. $$ \begin{array}{|l|l|} \hline x & P(X=x)\\ \hline 1 & \frac{1}{12}\\ \hline 2 & \frac{1}{12} \\ \hline 3 & \frac{1}{12} \\ \hline 4 & \frac{1}{6} \\ \hline 5 & \frac{1}{4} \\ \hline 6 & \frac{1}{3} \\ \hline \end{array} $$
Long-Term Probabilities of Our Crooked Die
| Empirical Probability | |||
| Roll Result |
Example
For our crooked die above, let $X$ be the possible value of each face. Calculate the following probabilities:
(a) $P(X=3)$
(b) $P(X<3)$
(c) $P(X \geq 3)$
(d) $P(3 \leq X \leq 5)$
Let's recall the PDF
$$
\begin{array}{|l|l|}
\hline
x & P(X=x)\\ \hline
1 & \frac{1}{12}\\ \hline
2 & \frac{1}{12} \\ \hline
3 & \frac{1}{12} \\ \hline
4 & \frac{1}{6} \\ \hline
5 & \frac{1}{4} \\ \hline
6 & \frac{1}{3} \\ \hline
\end{array}
$$
(a) $\displaystyle P(X=3)=\frac{1}{12}$
(b) $\displaystyle P(X \lt 3)=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$
(c) $\displaystyle P(X \geq 3)=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}+\frac{1}{3}=\frac{5}{6}$
Scenic Alternative
$\displaystyle P(X \geq 3)=1-P(X \lt 3)=1-\frac{1}{6}=\frac{5}{6}$
(d) $\displaystyle P(3 \leq X \leq 5)=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}=\frac{1}{2}$
(b) $\displaystyle P(X \lt 3)=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$
(c) $\displaystyle P(X \geq 3)=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}+\frac{1}{3}=\frac{5}{6}$
Scenic Alternative
$\displaystyle P(X \geq 3)=1-P(X \lt 3)=1-\frac{1}{6}=\frac{5}{6}$
(d) $\displaystyle P(3 \leq X \leq 5)=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}=\frac{1}{2}$
Example: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company.
Let $X$ = the number of years a new hire will stay with the company.
Let $P(X=x)$ = the probability that a new hire will stay with the company $x$ years.
Over the years, they have established the following PDF. $$ \begin{array}{|l|l|} \hline x & P(X=x) \\ \hline 0 & 0.13 \\ \hline 1 & \\ \hline 2 & 0.31 \\ \hline 3 & 0.14 \\ \hline 4 & 0.11 \\ \hline 5 & 0.09 \\ \hline 6 & 0.05 \\ \hline\end{array} $$
Example (Continued) Using the PDF provided above, answer the following:
(a) What is the probability a new hire will stay with the company for $1$ year?
(b) What is the probability a new hire will stay with the company for $1$ year or less?
(c) What is the probability a new hire will stay with the company for more than $1$ year?
(d) What is the probability a new hire will stay with the company between $1$ and $3$ years?
Recall the PDF.
$$
\begin{array}{|l|l|} \hline x & P(X=x) \\ \hline 0 & 0.13 \\ \hline 1 & \\ \hline 2 & 0.31 \\ \hline 3 & 0.14 \\ \hline 4 & 0.11 \\ \hline 5 & 0.09 \\ \hline 6 & 0.05 \\ \hline\end{array}
$$
(a) $P(X=1)=1-(0.13+0.31+0.14+0.11+0.09+0.05)=0.17$
(b) $P(X\leq 1)=0.13+0.17=0.30$
(c) $P(X\gt 1)=1-P(X\leq 1)=1-0.30=0.70$
(d) $P(1\leq X \leq 3)=0.17+00.31+0.14$
(b) $P(X\leq 1)=0.13+0.17=0.30$
(c) $P(X\gt 1)=1-P(X\leq 1)=1-0.30=0.70$
(d) $P(1\leq X \leq 3)=0.17+00.31+0.14$