### Probability Distribution

When we talk about the probability of discrete random variables, we normally talk about a probability distribution. In a **probability distribution**, you may have a table, a graph, or a chart that shows you all the possible values of \begin{align*}X\end{align*} (your variable), and the probability associated with each of these values \begin{align*}(P(X))\end{align*}.

It is important to remember that the values of a discrete random variable are not mutually inclusive. Think back to our car example with Jack and his mom. Jack could not, realistically, find a car that is both a Ford and a Mercedes (assuming he did not see a home-built car). He would either see a Ford or not see a Ford as he went from his car to the mall doors. Therefore, the values for the variable are mutually exclusive. Now let's look at an example.

#### Displaying a Probability Distribution

Say you are going to toss 2 coins. Show the probability distribution for this toss.

Let the variable \begin{align*}X\end{align*} be the number of times your coin lands on tails. The table below lists all of the possible events that can occur from the tosses.

Toss |
First Coin |
Second Coin |
\begin{align*}X\end{align*} |
---|---|---|---|

1 | H | H | 0 |

2 | H | T | 1 |

3 | T | T | 2 |

4 | T | H | 1 |

We can add a fifth column to the table above to show the probability of each of these events (the tossing of the 2 coins).

Toss |
First Coin |
Second Coin |
\begin{align*}X\end{align*} |
\begin{align*}P(X)\end{align*} |
---|---|---|---|---|

1 | H | H | 0 | \begin{align*}\frac{1}{4}\end{align*} |

2 | H | T | 1 | \begin{align*}\frac{1}{4}\end{align*} |

3 | T | T | 2 | \begin{align*}\frac{1}{4}\end{align*} |

4 | T | H | 1 | \begin{align*}\frac{1}{4}\end{align*} |

As you can see in the table, each event has an equally likely chance of occurring. You can see this by looking at the column \begin{align*}P(X)\end{align*}. From here, we can find the probability distribution. In the \begin{align*}X\end{align*} column, we have 3 possible discrete values for this variable: 0, 1, and 2.

\begin{align*}P(0) & = \text{toss} \ 1 = \frac{1}{4}\\ P(1) & = \text{toss} \ 2 + \text{toss} \ 4\\ & = \frac{1}{4} + \frac{1}{4}\\ & = \frac{1}{2}\\ P(2) & = \text{toss} \ 3 = \frac{1}{4}\end{align*}

#### Representing Probability Distributions Graphically

Represent the probability distribution from the previous example graphically.

Now we can represent the probability distribution with a graph, called a histogram. A **histogram** is a graph that uses bars vertically arranged to display data. Using the TI-84 PLUS calculator, we can draw the histogram to represent the data above. Let’s start by first adding the data into our lists. Below you will find the key sequence to perform this task. We will use this sequence frequently throughout the rest of this book, so make sure you follow along with your calculator.

This key sequence allows you to erase any data that may be entered into the lists already. Now let’s enter our data.

Now we can draw our histogram from the data we just entered.

The result is as follows:

We can see the values of \begin{align*}P(X)\end{align*} if we press \begin{align*}\boxed{\text{TRACE}}\end{align*}. Look at the screenshot below. You can see the value of \begin{align*}P(X) = 0.25\end{align*} for \begin{align*}X = 0\end{align*}.

It's clear that the histogram shows the probability distribution for the discrete random variable. In other words, \begin{align*}P(0)\end{align*} gives the probability that the discrete random variable is equal to 0, \begin{align*}P(1)\end{align*} gives the probability that the discrete random variable is equal to 1, and \begin{align*}P(2)\end{align*} gives the probability that the discrete random variable is equal to 2. Notice that the probabilities add up to 1. One of the rules for probability is that the sum of the probabilities of all the possible values of a discrete random variable must be equal to 1.

#### Identifying Probability Distributions

Does the following table represent the probability distribution for a discrete random variable?

\begin{align*}& X && 0 && 1 && 2 && 3\\ & P(X)&& 0.1 && 0.2 && 0.3 && 0.4\end{align*}

Yes, it does, since \begin{align*}\sum P(X)=0.1+0.2+0.3+0.4\end{align*}, or \begin{align*}\sum P(X)=1.0\end{align*}.

### Example

#### Example 1

Say you are going to spin a spinner 3 times and that the colors red and blue are equally represented. Show the probability distribution for these spins.

Let the variable \begin{align*}X\end{align*} be the number of times your spinner lands on red. The table below lists all of the possible events that can occur from the spins.

Trial |
First Spin |
Second Spin |
Third Spin |
\begin{align*}X\end{align*} |
---|---|---|---|---|

1 | R | R | R | 3 |

2 | R | R | B | 2 |

3 | R | B | R | 2 |

4 | R | B | B | 1 |

5 | B | R | R | 2 |

6 | B | B | R | 1 |

7 | B | R | B | 1 |

8 | B | B | B | 0 |

We can add a sixth column to the table above to show the probability of each of these events (the 3 spins of the spinner).

Trial |
First Spin |
Second Spin |
Third Spin |
\begin{align*}X\end{align*} |
\begin{align*}P(X)\end{align*} |
---|---|---|---|---|---|

1 | R | R | R | 3 | \begin{align*}\frac{1}{8}\end{align*} |

2 | R | R | B | 2 | \begin{align*}\frac{1}{8}\end{align*} |

3 | R | B | R | 2 | \begin{align*}\frac{1}{8}\end{align*} |

4 | R | B | B | 1 | \begin{align*}\frac{1}{8}\end{align*} |

5 | B | R | R | 2 | \begin{align*}\frac{1}{8}\end{align*} |

6 | B | B | R | 1 | \begin{align*}\frac{1}{8}\end{align*} |

7 | B | R | B | 1 | \begin{align*}\frac{1}{8}\end{align*} |

8 | B | B | B | 0 | \begin{align*}\frac{1}{8}\end{align*} |

As you can see in the table, each event has an equally likely chance of occurring. You can see this by looking at the column \begin{align*}P(X)\end{align*}. From here, we can find the probability distribution. In the \begin{align*}X\end{align*} column, we have 4 possible discrete values for this variable: 0, 1, 2, and 3.

\begin{align*}P(0) & = \text{trial} \ 8 = \frac{1}{8}\\ P(1) & = \text{trial} \ 4 + \text{trial} \ 6 + \text{trial} \ 7\\ & = \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\\ & = \frac{3}{8}\\ P(2) & = \text{trial} \ 2 + \text{trial} \ 3 + \text{trial} \ 5\\ & = \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\\ & = \frac{3}{8}\\ P(3) & = \text{trial} \ 1 = \frac{1}{8}\end{align*}

### Review

- Does the following table represent the probability distribution for a discrete random variable? \begin{align*}& X && 2 && 4 && 6 && 8\\ & P(X) && 0.2 && 0.4 && 0.6 && 0.8\end{align*}
- Does the following table represent the probability distribution for a discrete random variable? \begin{align*}& X && 1 && 2 && 3 && 4 && 5\\ & P(X) && 0.202 && 0.174 && 0.096 && 0.078 && 0.055\end{align*}
- Does the following table represent the probability distribution for a discrete random variable? \begin{align*}& X && 1 && 2 && 3 && 4 && 5 && 6\\ & P(X) && 0.302 && 0.251 && 0.174 && 0.109 && 0.097 && 0.067\end{align*}

Say you are going to spin a spinner 2 times and that the colors green, yellow, and purple are equally represented. Let the variable \begin{align*}X\end{align*} be the number of times your spinner lands on green. The table below lists all of the possible events that can occur from the spins.

Trial |
First Spin |
Second Spin |
\begin{align*}X\end{align*} |
---|---|---|---|

1 | G | G | 2 |

2 | G | Y | 1 |

3 | G | P | 1 |

4 | Y | G | 1 |

5 | Y | Y | 0 |

6 | Y | P | 0 |

7 | P | G | 1 |

8 | P | Y | 0 |

9 | P | P | 0 |

- What is the probability that the spinner doesn't land on green on either of the spins?
- What is the probability that the spinner lands on green on 1 of the spins?
- What is the probability that the spinner lands on green on both of the spins?
- Create the histogram for this scenario on your TI calculator.
- Use the TRACE feature of your TI calculator to show the probability that the spinner doesn't land on green on either of the spins.
- Use the TRACE feature of your TI calculator to show the probability that the spinner lands on green on 1 of the spins.
- Use the TRACE feature of your TI calculator to show the probability that the spinner lands on green on both of the spins.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 3.2.