3.1: Events, Sample Spaces, and Probability

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Know basic statistical terminology.
• List simple events and sample spaces.
• Know the basic rules of probability.

Introduction

The concept of probability plays an important role in our daily lives. Assume you have an opportunity to invest some money in a software company. Suppose you know that the company’s records indicate that in the past five years, its profits have been consistently decreasing. Would you still invest your money in it? Do you think the chances are good for the company in the future?

Here is another illustration. Suppose that you are playing a game that involves tossing a single die. Assume that you have already tossed it 10 times, and every time the outcome was the same, a 2. What is your prediction of the eleventh toss? Would you be willing to bet $100 that you will not get a 2 on the next toss? Do you think the die is loaded? Notice that the decision concerning a successful investment in the software company and the decision of whether or not to bet$100 on the next outcome of the die are both based on probabilities of certain sample results. Namely, the software company’s profits have been declining for the past five years, and the outcome of rolling a 2 ten times in a row seems strange. From these sample results, we might conclude that we are not going to invest our money in the software company or bet on this die. In this lesson, you will learn mathematical ideas and tools that can help you understand such situations.

Events, Sample Spaces, and Probability

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, H\begin{align*}H\end{align*}, or tails, T\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.

Example: 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.

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

HTHHHTHTHTTT\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 S\begin{align*}S\end{align*}.

Example: We want to determine the sample space of throwing a die and the sample space of tossing a coin.

Solution: 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:

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

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

Example: 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, R\begin{align*}R\end{align*}, on the first drawing and then another red one on the second, RR\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:

S={RR,RB,RW,BR,BB,BW,WR,WB,WW}\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, RW\begin{align*}RW\end{align*} is different from WR\begin{align*}WR\end{align*}.

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

RBWRRRBRWRBRBBBWBWRWBWWW\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.

Example: 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:

S={RB,RW,BR,BW,WR,WB}\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, T\begin{align*}T\end{align*}, is the same as the chance of getting heads, H\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, P\begin{align*}P\end{align*}, of an outcome, A\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: 0P(A)1\begin{align*}0 \le P(A) \le 1\end{align*}, for any event, A\begin{align*}A\end{align*}.

The probability of an event, A\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 0.5+0.5=1.\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: P(A)=1\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, 16\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 P\begin{align*}P\end{align*}, and the respective elements of the sample space (the outcomes) are denoted by A,B,C,\begin{align*}A, B, C,\end{align*} etc. The mathematical notation that indicates the probability that an outcome, A\begin{align*}A\end{align*}, happens is P(A)\begin{align*}P(A)\end{align*}. We use the following formula to calculate the probability of an outcome occurring:

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

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

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

P(A)=The number of outcomes for HH to occurThe size of the sample space=14=25%\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.

Example: What is the probability of throwing a die and getting A=2,3, or 4\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.

P(A)=The number of outcomes for 2, 3, or 4 to occurThe size of the sample space=36=12=50%\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.

Example: Consider tossing 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
HH\begin{align*}HH\end{align*} 49\begin{align*}\frac{4}{9}\end{align*}
HT\begin{align*}HT\end{align*} 29\begin{align*}\frac{2}{9}\end{align*}
TH\begin{align*}TH\end{align*} 29\begin{align*}\frac{2}{9}\end{align*}
TT\begin{align*}TT\end{align*} 19\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 HT\begin{align*}HT\end{align*} and TH\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*}

Lesson Summary

An event is something that occurs, or happens, with one or more possible outcomes.

An experiment is the process of taking a measurement or making an observation.

A simple event is the simplest outcome of an experiment.

The sample space is the set of all possible outcomes of an experiment, typically denoted by \begin{align*}S\end{align*}.

For a description of how to find an event given a sample space (1.0), see teachertubemath, Probability Events (2:23).

Review Questions

1. Consider an experiment composed of throwing a die followed by throwing a coin.
1. List the simple events and assign a probability for each simple event.
2. What are the probabilities of observing the following events?

(i) A \begin{align*}2\end{align*} on the die and \begin{align*}H\end{align*} on the coin

(ii) An even number on the die and \begin{align*}T\end{align*} on the coin

(iii) An even number on the die

(iv) \begin{align*}T\end{align*} on the coin

1. 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*}
1. Find \begin{align*}P(A)\end{align*}
2. Find \begin{align*}P(B)\end{align*}
2. 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*}.
1. List the outcomes in the sample space.
2. Determine the probability of observing each of the following events:

(i) Drawing 2 blue marbles

(ii) Drawing 1 red marble and 1 blue marble

(iii) Drawing 2 red marbles

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