### Basic Terminology and Probability Rules

An **event** is something that occurs, or happens. For example, flipping a coin is an event, and so is walking in the park and passing by a bench. Anything that could possibly happen is an event.

Every event has one or more possible outcomes. While tossing a coin is an event, getting tails is the outcome of that event. Likewise, while walking in the park is an event, finding your friend sitting on the bench is an outcome of that event.

Suppose a coin is tossed once. There are two possible outcomes, either heads, \begin{align*}H\end{align*}, or tails, \begin{align*}T\end{align*}. Notice that if the experiment is conducted only once, you will observe only one of the two possible outcomes. An **experiment** is the process of taking a measurement or making an observation. These individual outcomes for an experiment are each called **simple events**.

A die has six possible outcomes: 1, 2, 3, 4, 5, or 6. When we toss it once, only one of the six outcomes of this experiment will occur. The one that does occur is called a simple event.

Suppose that two pennies are tossed simultaneously. We could have both pennies land heads up (which we write as \begin{align*}HH\end{align*}), or the first penny could land heads up and the second one tails up (which we write as \begin{align*}HT\end{align*}), etc. We will see that there are four possible outcomes for each toss, which are \begin{align*}HH, HT, TH\end{align*}, and \begin{align*}TT\end{align*}. The table below shows all the possible outcomes.

\begin{align*}&& H && T\\ H && HH && HT\\ T && TH && TT\end{align*}

**Figure:** The possible outcomes of flipping two coins.

What we have accomplished so far is a listing of all the possible simple events of an experiment. This collection is called the **sample space** of the experiment.

The sample space is the set of all possible outcomes of an experiment, or the collection of all the possible simple events of an experiment. We will denote a sample space by \begin{align*}S\end{align*}.

As we know, there are 6 possible outcomes for throwing a die. We may get 1, 2, 3, 4, 5, or 6, so we write the sample space as the set of all possible outcomes:

\begin{align*}S = \left \{1, 2, 3, 4, 5, 6 \right \}\end{align*}

Similarly, the sample space of tossing a coin is either heads, \begin{align*}H\end{align*}, or tails, \begin{align*}T\end{align*}, so we write \begin{align*}S=\left \{H,T\right \}\end{align*}.

#### Determining the Sample Space

Suppose a box contains three balls, one red, one blue, and one white. One ball is selected, its color is observed, and then the ball is placed back in the box. The balls are scrambled, and again, a ball is selected and its color is observed. What is the sample space of the experiment?

It is probably best if we draw a **tree diagram** to illustrate all the possible selections.

As you can see from the tree diagram, it is possible that you will get the red ball, \begin{align*}R\end{align*}, on the first drawing and then another red one on the second, \begin{align*}RR\end{align*}. You can also get a red one on the first and a blue on the second, and so on. From the tree diagram above, we can see that the sample space is as follows:

\begin{align*}S = \left \{RR, RB, RW, BR, BB, BW, WR, WB, WW \right \}\end{align*}

Each pair in the set above gives the first and second drawings, respectively. That is, \begin{align*}RW\end{align*} is different from \begin{align*}WR\end{align*}.

We can also represent all the possible drawings by a table or a matrix:

\begin{align*}&& R && B && W\\ R && RR && RB && RW\\ B && BR && BB && BW\\ W && WR && WB && WW\end{align*}

**Figure:** Table representing the possible outcomes diagrammed in the previous figure. The first column represents the first drawing, and the first row represents the second drawing.

Consider the same experiment as in the last example. This time we will draw one ball and record its color, but we will not place it back into the box. We will then select another ball from the box and record its color. What is the sample space in this case?

Solution: The tree diagram below illustrates this case:

You can clearly see that when we draw, say, a red ball, the blue and white balls will remain. So on the second selection, we will either get a blue or a while ball. The sample space in this case is as shown:

\begin{align*}S= \left \{RB, RW, BR, BW, WR, WB\right \}\end{align*}

Now let us return to the concept of probability and relate it to the concepts of sample space and simple events. If you toss a fair coin, the chance of getting tails, \begin{align*}T\end{align*}, is the same as the chance of getting heads, \begin{align*}H\end{align*}. Thus, we say that the probability of observing heads is 0.5, and the probability of observing tails is also 0.5. The probability, \begin{align*}P\end{align*}, of an outcome, \begin{align*}A\end{align*}, always falls somewhere between two extremes: 0, which means the outcome is an impossible event, and 1, which means the outcome is guaranteed to happen. Most outcomes have probabilities somewhere in-between.

Property 1: \begin{align*}0 \le P(A) \le 1\end{align*}, for any event, \begin{align*}A\end{align*}.

The probability of an event, \begin{align*}A\end{align*}, ranges from 0 (impossible) to 1 (certain).

In addition, the probabilities of all possible simple outcomes of an event must add up to 1. This 1 represents certainty that one of the outcomes must happen. For example, tossing a coin will produce either heads or tails. Each of these two outcomes has a probability of 0.5. This means that the total probability of the coin landing either heads or tails is \begin{align*}0.5 + 0.5 = 1.\end{align*} That is, we know that if we toss a coin, we are certain to get heads or tails.

Property 2: \begin{align*}\sum P(A)=1\end{align*} when summed over all possible simple outcomes.

The sum of the probabilities of all possible outcomes must add up to 1.

Notice that tossing a coin or throwing a die results in outcomes that are all equally probable. That is, each outcome has the same probability as all the other outcomes in the same sample space. Getting heads or tails when tossing a coin produces an equal probability for each outcome, 0.5. Throwing a die has 6 possible outcomes, each also having the same probability, \begin{align*}\frac{1}{6}\end{align*}. We refer to this kind of probability as classical probability. *Classical probability* is defined to be the ratio of the number of cases favorable to an event to the number of all outcomes possible, where each of the outcomes is equally likely.

Probability is usually denoted by \begin{align*}P\end{align*}, and the respective elements of the sample space (the outcomes) are denoted by \begin{align*}A, B, C,\end{align*} etc. The mathematical notation that indicates the probability that an outcome, \begin{align*}A\end{align*}, happens is \begin{align*}P(A)\end{align*}. We use the following formula to calculate the probability of an outcome occurring:

\begin{align*}P(A)=\frac{\text{The number of outcomes for} \ A \ \text{to occur}}{\text{The size of the sample space}}\end{align*}

#### Determining Probability

When tossing two coins, what is the probability of getting a head on both coins, \begin{align*}HH\end{align*}? Is the probability classical?

Since there are 4 elements (outcomes) in the sample space set, \begin{align*}\left \{HH, HT, TH, TT\right \}\end{align*}, its size is 4. Furthermore, there is only 1 \begin{align*}HH\end{align*} outcome that can occur. Therefore, using the formula above, we can calculate the probability as shown:

\begin{align*}P(A)=\frac{\text{The number of outcomes for} \ HH \ \text{to occur}}{\text{The size of the sample space}}=\frac{1}{4}=25\%\end{align*}

Notice that each of the 4 possible outcomes is equally likely. The probability of each is 0.25. Also notice that the total probability of all possible outcomes in the sample space is 1.

What is the probability of throwing a die and getting \begin{align*}A = 2, 3, \ \text{or} \ 4\end{align*}?

There are 6 possible outcomes when you toss a die. Thus, the total number of outcomes in the sample space is 6. The event we are interested in is getting a 2, 3, or 4, and there are three ways for this event to occur.

\begin{align*}P(A)=\frac{\text{The number of outcomes for 2, 3, or 4 to occur}}{\text{The size of the sample space}}=\frac{3}{6}=\frac{1}{2}=50\%\end{align*}

Therefore, there is a probability of 0.5 that we will get 2, 3, or 4.

#### Using a Probability Table

Suppose you toss two coins. Assume the coins are not balanced. The design of the coins is such that they produce the probabilities shown in the table below:

Outcome |
Probability |
---|---|

\begin{align*}HH\end{align*} | \begin{align*}\frac{4}{9}\end{align*} |

\begin{align*}HT\end{align*} | \begin{align*}\frac{2}{9}\end{align*} |

\begin{align*}TH\end{align*} | \begin{align*}\frac{2}{9}\end{align*} |

\begin{align*}TT\end{align*} | \begin{align*}\frac{1}{9}\end{align*} |

**Figure:** Probability table for flipping two weighted coins.

What is the probability of observing exactly one head, and what is the probability of observing at least one head?

Notice that the simple events \begin{align*}HT\end{align*} and \begin{align*}TH\end{align*} each contain only one head. Thus, we can easily calculate the probability of observing exactly one head by simply adding the probabilities of the two simple events:

\begin{align*}P & = P(HT)+P(TH)\\ & =\frac{2}{9}+\frac{2}{9}\\ & =\frac{4}{9}\end{align*}

Similarly, the probability of observing at least one head is:

\begin{align*}P& =P(HH)+P(HT)+P(TH)\\ & =\frac{4}{9}+\frac{2}{9}+\frac{2}{9}=\frac{8}{9}\end{align*}

### Examples

Suppose you have a jar of candies: 4 red, 5 purple and 7 green. Find the probability of the following events.

#### Example 1

Selecting a red candy

To find the probability of selecting a red candy, first find the total number of candies:

\begin{align*}total=4+5+7=16\end{align*}

Since there are 4 ways to get a red candy out of a total of 16 candies, the probability of selecting a red candy is:

\begin{align*}P(R)=\frac{4}{16}=\frac{1}{4}=0.25\end{align*}

#### Example 2

Selecting a purple candy.

To find the probability of selecting a purple candy, since there are 5 ways to get a purple candy out of a total of 16 candies, the probability of selecting a purple candy is:

\begin{align*}P(P)=\frac{5}{16}= 0.3125\end{align*}

#### Example 3

Selecting a green candy.

To find the probability of selecting a green candy, since there are 7 ways to get a green candy out of a total of 16 candies, the probability of selecting a green candy is:

\begin{align*}P(G)=\frac{7}{16}= 0.4375\end{align*}

#### Example 4

Selecting a yellow candy.

Since there are no yellow candies in the jar, the probability of selecting a yellow candy is 0.

### Review

For 1-4, consider an experiment composed of throwing a die followed by throwing a coin.

a. List the simple events and assign a probability for each simple event.

b. What are the probabilities of observing the following events?

- A \begin{align*}2\end{align*} on the die and \begin{align*}H\end{align*} on the coin.
- A \begin{align*}2\end{align*} on the die and \begin{align*}H\end{align*} on the coin.
- An even number on the die.
- \begin{align*}T\end{align*} on the coin.

For 5-6, the Venn diagram below shows an experiment with six simple events. Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are also shown. The probabilities of the simple events

are:

\begin{align*}P(1)& =P(2)=P(4)=\frac{2}{9}\\ P(3)& =P(5)=P(6)=\frac{1}{9}\end{align*}

- Find \begin{align*}P(A)\end{align*}
- Find \begin{align*}P(B)\end{align*}

For 7-10, a box contains two blue marbles and three red ones. Two marbles are drawn randomly without replacement. Refer to the blue marbles as \begin{align*}B1\end{align*} and \begin{align*}B2\end{align*} and the red ones as \begin{align*}R1\end{align*}, \begin{align*}R2\end{align*}, and \begin{align*}R3\end{align*}.

a. List the outcomes in the sample space.

b. Determine the probability of observing each of the event.

- Drawing 2 blue marbles.
- Drawing 1 red marble and 1 blue marble.
- Drawing 2 red marbles.
- Neither marble is blue or red.

- List the sample space for the following:
- Spinning a round spinner labeled A, B, C, D, E
- The sexes of a 3-child family
- The order in which 4 blocks A, B, C, and D can be lined up

- Show on a 2-dimensional grid the sample space for
- Rolling a die and tossing a coin at the same tie.
- Tossing two coins
- Rolling a die and spinning a spinner with sides A, B, C and D

- Show on a tree diagram the sample space for:
- Tossing a nickel and a dime at the same time.
- Tossing a coin three times
- Drawing two tickets from a hat containing a number of pink, blue and white tickets.

- Is each of the following values a legitimate probability value? Explain any “no” answers.
- .56
- .00
- 1.00
- 1.43
- -.36

- An appliance storeowner notices that one out of 30 customers returns the appliance within the two weeks of making the purchase.
- Write a sentence expressing this fact as a proportion.
- Write a sentence expressing this fact as a percent.
- Write a sentence expressing this fact as a probability.

- For each of the following experiments, describe the sample space.
- Toss a coin four times.
- Measure the lifetime (in hours) of a particular brand of light bulb.
- Record the weights of 13 day old rats.
- Observe the proportion of defectives in a shipment of computer components.

### Review (Answers)

To view the review answers, open this PDF file and look for section 3.1.