# 3.1: Discrete Random Variables

**At Grade**Created by: CK-12

**Learning Objectives**

- Demonstrate an understanding of the notion of discrete random variables by using them to solve for the probabilities of outcomes, such as the probability of the occurrence of five heads in 14 coin tosses.

You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins.

(a) Determine the theoretical probability for the teacher.

(b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials.

Work through Chapter 3 and then revisit this problem to find the solution.

Whenever you run and experiment, flip a coin, roll a die, pick a card, you assign a number to represent the value to the outcome that you get. This number that you assign is called a **random variable**. For example, if you were to roll two dice and asked what the sum of the two dice might be, you would design the following table of numerical values.

These numerical values represent the possible outcomes of the rolling of two dice and summing of the result. In other words, rolling one die and seeing a \begin{align*}{\color{red}6}\end{align*}

The rolling of a die is interesting because there are only a certain number of possible outcomes that you can get when you roll a typical die. In other words, a typical die has the numbers 1, 2, 3, 4, 5, and 6 on it and nothing else. A **discrete random variable** can only have a specific (or finite) number of numerical values.

A random variable is simply the rule that assigns the number to the outcome. For our example above, there are 36 possible combinations of the two dice being rolled. The discrete random variables (or values) in our sample are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, as you can see in the table below.

We can have infinite discrete random variables if we think about things that we know have an estimated number. Think about the number of stars in the universe. We know that there are not a specific number that we have a way to count so this is an example of an infinite discrete random variable. Another example would be with investments. If you were to invest $1000 at the start of this year, you could only estimate the amount you would have at the end of this year.

Well, how does this relate to probability?

** Example 1:** Looking at the previous table, what is the probability that the sum of the two dice rolled would be 4?

*Solution:*

\begin{align*}P(4) & =\frac{3}{36} \\
P(4) & =\frac{1}{12}\end{align*}

** Example 2:** A coin is tossed 3 times. What are the possible outcomes? What is the probability of getting one head?

*Solution:*

If our first toss were a heads...

If our first toss were a tails...

Therefore the possible outcomes are:

\begin{align*}&{\color{blue}HHH}, \ {\color{blue}HH}T, \ {\color{blue}H}T{\color{blue}H}, \ {\color{blue}H}TT, \ T{\color{blue}HH}, \ T{\color{blue}H}T, \ TT{\color{blue}H}, TTT\\
& P(1\ {\color{blue}\text{head}}) = \frac{3}{8}\end{align*}

*Alternate Solution:*

We have one coin and want to find the probability of getting one head in three tosses. We need to calculate two parts to solve the probability problem.

*Numerator (Top)*

In our example, we want to have 1 H and 2Ts. Our favorable outcomes would be any combination of HTT. The number of favorable choices would be:

\begin{align*}\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\
\# \ of\ favorable\ choices & = \frac{3\ letters!}{1\ head! \times 2\ tails!} \\
\# \ of\ favorable\ choices & = \frac{3 \times 2 \times 1}{1 \times (2 \times 1)} \\
\# \ of\ favorable\ choices & = \frac{6}{2} = 3\end{align*}

*Denominator (Bottom)*

The number of possible outcomes \begin{align*}= 2 \times 2 \times 2 = 8\end{align*}

We now want to find the number of possible times we could get one head when we do these three tosses. We call these favorable outcomes. Why? Because these are the outcomes that we want to happen, therefore they are favorable.

Now we just divide the numerator by the denominator.

\begin{align*}P(1\ head) = \frac{3}{8}\end{align*}

**Remember:**

Possible outcomes \begin{align*}= 2^n\end{align*}**number of tosses.**

Here we have \begin{align*}{\color{red}3}\end{align*}

Possible outcomes \begin{align*}= 2^{\color{red}n}\end{align*}

Possible outcomes \begin{align*}= 2^{\color{red}3}\end{align*}

Possible outcomes \begin{align*}= 2 \times 2 \times 2\end{align*}

Possible outcomes \begin{align*}= 8\end{align*}

**Note:** The **factorial function** **(symbol: !)** just means to multiply a series of descending natural numbers.

Examples:

\begin{align*}{\color{blue}4!} & {\color{blue}\ = 4 \times 3 \times 2 \times 1 = 24}\\
{\color{blue}7!} & {\color{blue}\ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040} \\
{\color{blue}1!} & {\color{blue}\ = 1}\end{align*}

Note: It is generally agreed that \begin{align*}0! = 1\end{align*}

** Example 3:** A coin is tossed 4 times. What are the possible outcomes? What is the probability of getting one head?

*Solution:*

If our first toss were a heads...

If our first toss were a tails...

Therefore there are 16 possible outcomes:

\begin{align*}&{\color{blue}HHHH}, \ {\color{blue}HHH}T, \ {\color{blue}HH}T{\color{blue}H}, \ {\color{blue}HH}TT, \ {\color{blue}H}T{\color{blue}HH}, \ {\color{blue}H}T{\color{blue}H}T, \ {\color{blue}H}TT{\color{blue}H}, \ {\color{blue}H}TTT, \ T{\color{blue}HHH},\\
& \ T{\color{blue}HH}T, \ T{\color{blue}H}T{\color{blue}H}, \ T{\color{blue}H}TT, \ TT{\color{blue}HH}, \ TT{\color{blue}H}T, \ TTT{\color{blue}H}, \ TTTT \\
& P(1\ {\color{blue}\text{head}}) = \frac{4}{16} \\
& P(1\ {\color{blue}\text{head}}) = \frac{1}{4}\end{align*}

*Alternate Solution:*

We have one coin and want to find the probability of getting one head in four tosses. We need to calculate two parts to solve the probability problem.

*Numerator (Top)*

In our example, we want to have 1 H and 3 Ts. Our favorable outcomes would be any combination of HTTT. The number of favorable choices would be:

\begin{align*}\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\
\# \ of\ favorable\ choices & = \frac{4\ letters!}{1\ head! \times 3\ tails!} \\
\# \ of\ favorable\ choices & = \frac{4 \times 3 \times 2 \times 1}{1 \times (3 \times 2 \times 1)} \\
\# \ of\ favorable\ choices & = \frac{24}{6} \\
\# \ of\ favorable\ choices & = 4\end{align*}

*Denominator (Bottom)*

The number of possible outcomes \begin{align*}= 2 \times 2 \times 2 \times 2 = 16\end{align*}

We now want to find the number of possible times we could get one head when we do these four tosses (or our favorable outcomes).

**Remember:**

Possible outcomes \begin{align*}= 2^n\end{align*}**number of tosses.**

Here we have \begin{align*}{\color{red}4}\end{align*}

Possible outcomes \begin{align*}= 2^{{\color{red}n}}\end{align*}

Possible outcomes \begin{align*}= 2^{{\color{red}4}}\end{align*}

Possible outcomes \begin{align*}= 2 \times 2 \times 2 \times 2\end{align*}

Possible outcomes \begin{align*}= 16\end{align*}

Now we just divide the numerator by the denominator.

\begin{align*}P(1\ head) & = \frac{4}{16} \\
P(1\ head) & = \frac{1}{4}\end{align*}

**Technology Note:**

Let’s take a look at how we can do this using the TI-84 calculators. There is an application on the TI calculators called the coin toss. Among others (including the dice roll, spinners, and picking random numbers), the coin toss is an excellent application for when you what to find the probabilities for a coin tossed more than 4 times or more than one coin being tossed multiple times.

Let’s say you want to see one coin being tossed one time. Here is what the calculator will show and the key strokes to get to this toss.

Let’s say you want to see one coin being tossed ten times. Here is what the calculator will show and the key strokes to get to this sequence. Try it on your own.

We can actually see how many heads and tails occurred in the tossing of the 10 coins. If you click on the right arrow (>) the frequency label will show you how many of the tosses came up heads.

We could also use ** randBin** to simulate the tossing of a coin. Follow the keystrokes below.

This list contains the count of heads resulting from each set of 10 coin tosses. If you use the right arrow (>) you can see how many times from the 20 trials you actually had 4 heads.

Now let’s go back to our original chapter problem and see if we have gained enough knowledge to answer it.

You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins.

(a) Determine the theoretical probability for the teacher.

(b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials.

*Solution:*

(a) Let’s calculate the theoretical probability of getting 5 heads for the 14 tosses.

*Numerator (Top)*

In our example, we want to have 5 H and 9 Ts. Our favorable outcomes would be any combination of HHHHHTTTTTTTTT. The number of favorable choices would be:

\begin{align*}\#\ of\ favorable\ choices & = \frac{\#\ possible\ letters\ in\ combination!}{letter\ X! \times letter\ Y!} \\
\# \ of\ favorable\ choices & = \frac{14\ letters!}{5\ head! \times 9\ tails!} \\
\# \ of\ favorable\ choices & = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)\times (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\
\# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(120) \times (362880)} \\
\# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(43545600)} \\
\# \ of\ favorable\ choices & = 2002\end{align*}

*Denominator (Bottom)*

The number of possible outcomes \begin{align*}= 2^{14}\end{align*}

The number of possible outcomes \begin{align*}= 16384\end{align*}

Now we just divide the numerator by the denominator.

\begin{align*}P(5\ heads) & = \frac{2002}{16384} \\ P(5\ heads) & = 0.1222\end{align*}

The probability would be 12% of the tosses would have 5 heads.

b)

Looking at the data that resulted in this trial, there were 4 times of 20 that 5 heads appeared.

\begin{align*}P(5\ \text{heads}) = \frac{4}{20}\end{align*} or 20%.

**Lesson Summary**

Probability in this chapter focused on experiments with random variables or the numbers that you assign to the probability of events. If we have a discrete random variable, then there are only a specific number of variables we can choose from. For example, tossing a fair coin has a probability of success for heads = probability of success for tails \begin{align*}= 0.50\end{align*}. Using tree diagrams or the formula \begin{align*}P = \frac{\#\ of\ favorable\ outcomes}{total\ \#\ of\ outcomes}\end{align*}, we can calculate the probabilities of these events. Using the formula requires the use of the factorial function where numbers are multiplied in descending order.

**Points to Consider**

- How is the calculator a useful tool for calculating probability in discrete random variable experiments?
- Are TREE Diagrams useful in interpreting the probability of simple events?

**Vocabulary**

- Discrete Random Variables
- Only have a specific (or finite) number of numerical values.

- Random Variable
- A variable that takes on numerical values governed by a chance experiment.

- Factorial Function
- (symbol: !) – The function of multiplying a series of descending natural numbers.

- Theoretical Probability
- A probability calculated by analyzing a situation, rather than performing an experiment, given by the ratio of the number of different ways an event can occur to the total number of equally likely outcomes possible. The numerical measure of the likelihood that an event, \begin{align*}E\end{align*}, will happen.

\begin{align*}P(E) = \frac{number\ of\ favorable\ outcomes}{total\ number\ of\ possible\ outcomes}\end{align*}

- Tree Diagram
- A branching diagram used to list all the possible outcomes of a compound event.

**Review Questions**

- Define and give three examples of discrete random variables.
- Draw a tree diagram to represent the tossing of two coins and determine the probability of getting at least one head.
- Draw a tree diagram to represent the tossing of one coin three times and determine the probability of getting at least one head.
- Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at least one red.
- Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at two blue marbles. \begin{align*}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\\ & P(\text{two blue marbles}) =\frac{1}{9}\end{align*}
- Draw a diagram to represent the rolling two dice and determine the probability of getting at least one 5.
- Draw a diagram to represent the rolling two dice and determine the probability of getting two 5s.
- Use
**randBin**to simulate the 6 tosses of a coin 20 times to determine the probability of getting two tails. - Use
**randBin**to simulate the 15 tosses of a coin 25 times to determine the probability of getting two heads. - Calculate the theoretical probability of getting 4 heads for the 12 tosses.
- Calculate the theoretical probability of getting 8 heads for the 10 tosses. \begin{align*}P(8 \ heads) = \frac{45}{1024}\end{align*}
- Calculate the theoretical probability of getting 8 heads for the 15 tosses.

**Review Answers**

- Answers will vary
- \begin{align*}P(\text{at least}\ I \ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\ P(\text{at least}\ I \ H) & = \frac{3}{4}\end{align*}
- \begin{align*}P(\text{at least}\ I\ H) & = \frac{HHH,HHT,HTH,HTT,THH,THT,TTH}{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} \\ P(\text{at least}\ I\ H) & = \frac{7}{8}\end{align*}
- \begin{align*}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\end{align*} \begin{align*}P(\text{at least one red}) &= \frac{3}{9} \\ P(\text{at least one red}) &= \frac{1}{3}\end{align*}
- .
- \begin{align*}P (\text{at least one}\ 5) = \frac{11}{36}\end{align*}
- \begin{align*}P\end{align*}(two 5's) \begin{align*}=\frac{1}{36}\end{align*}
- \begin{align*}P (2 \ \text{heads}) = \frac{4}{20} = 20\%\end{align*}
- \begin{align*}P\end{align*}(4 heads) \begin{align*}= \frac{6}{25} = 24\%\end{align*}
- \begin{align*}\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{967680} \\ \# \ of\ favorable\ choices & = 495\end{align*} The number of possible outcomes \begin{align*}= 2^{12}\end{align*} The number of possible outcomes \begin{align*}= 4096\end{align*} Now we just divide the numerator by the denominator. \begin{align*}P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%\end{align*}
- \begin{align*}P(8\ \text{heads}) = 4.39\%\end{align*}
- \begin{align*}\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\ \# \ of\ favorable\ choices & = 6446\end{align*} The number of possible outcomes \begin{align*}= 2^{15}\end{align*} The number of possible outcomes \begin{align*}= 32768\end{align*} Now we just divide the numerator by the denominator. \begin{align*}P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%\end{align*}

**Answer Key for Review Questions (even numbers)**

2.

\begin{align*}P(\text{at least}\ I\ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\ P(\text{at least}\ I\ H) & = \frac{3}{4}\end{align*}

4.

\begin{align*}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\\ & P(\text{at least one red}) = \frac{3}{9} \\ & P(\text{at least one red}) = \frac{1}{3}\end{align*}

6.

\begin{align*}P\end{align*} (at least one 5) \begin{align*}= \frac{11}{36}\end{align*}

8.

\begin{align*}P\end{align*} (2 heads) \begin{align*}= \frac{4}{20} = 20\%\end{align*}

10. \begin{align*}\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{967680} \\ \# \ of\ favorable\ choices & = 495\end{align*}

The number of possible outcomes \begin{align*}= 2^{12}\end{align*}

The number of possible outcomes \begin{align*}= 4096\end{align*}

Now we just divide the numerator by the denominator.

\begin{align*}P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%\end{align*}

12. \begin{align*}\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\ \# \ of\ favorable\ choices & = 6446\end{align*}

The number of possible outcomes \begin{align*}= 2^{15}\end{align*}

The number of possible outcomes \begin{align*}= 32768\end{align*}

Now we just divide the numerator by the denominator.

\begin{align*}P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%\end{align*}

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