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3.1: Events, Sample Spaces, and Probability

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

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

An event is something that occurs or happens. Flipping a coin is an event. Walking in the park and passing by a bench is an event. Anything that could possibly happen is an event.

Every event has one or more possible outcomes. Tossing a coin is an event but getting a tail is the outcome of the event. Walking in the park is an event and finding your friend sitting on a bench is an outcome of the event.

In statistics, the process of taking a measurement or making an observation is called an experiment. For example, tossing a coin and recording the up face in a table of data is an experiment because a measurement is taken.


The process of taking a measurement or making an observation.

Keep in mind that the definition of an experiment in statistics is broader than the one used in science. A scientific experiment involves scientific instrumentations such as thermometers, microscopes, telescopes and tubes. A statistical experiment may involve all these items but it mainly involves recording data and measurements. For example, we may conduct an experiment to learn which brand of coffee a customer may prefer among three brands, recording a voter’s opinion on a particular political issue, or measuring the amount of carbon monoxide present in a certain environment. Any kind of observation, measuring, and recording that you may conduct can be considered a statistical experiment.

Suppose a coin is tossed once. There are two possible outcomes, either a head \begin{align*}(H)\end{align*} or a tail \begin{align*}(T)\end{align*}. Notice that if the experiment is conducted only once, you will observe only one of the two possible outcomes. These individual outcomes for an experiment are each called simple events. Here is another example: a die has six possible outcomes: \begin{align*}1, 2, 3, 4, 5, 6.\end{align*} 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.

Simple Event

The simplest outcome of an experiment.


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. In other words, the simple events 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 an experiment.

Sample Space

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*}.


Experiment: We want to investigate the sample space of throwing a die and the sample space of tossing a coin.


As we know, there are \begin{align*}6\end{align*} possible outcomes for throwing a die. We may get \begin{align*}1, 2, 3, 4, 5,\end{align*} or \begin{align*}6\end{align*}. 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 head \begin{align*}(H)\end{align*} or tail \begin{align*}(T)\end{align*} so we write \begin{align*}S = \left \{ H, T \right \}\end{align*}.


Experiment: 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 diagram to illustrate all the possible drawings.

As you can see from the 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 diagram above, we can see that the sample space is:

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

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

Where the first column represents the first drawing and the first row represents the second drawing.


Experiment: Consider the same experiment as in the example before last but 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?


The diagram below illustrates this case:

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

\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 that we have just studied. You may be familiar with the meaning of probability and may have used the term as a synonym with informal words like “chance” and “odds.” For the time being, we will begin our treatment of probability using these informal concepts and then later, we will solidify these meanings into formal mathematical definitions.

As you probably know from your previous math courses, if you toss a fair coin, the chance of getting a tail \begin{align*}(T)\end{align*} is the same as the chance of getting a head \begin{align*}(H)\end{align*}. Thus we say that the probability of observing a head is \begin{align*}50 \%\end{align*} (or \begin{align*}0.5\end{align*}) and the probability of observing a tail is also \begin{align*}50 \%\end{align*}. We also say sometimes “the odds are \begin{align*}50-50\end{align*}.”

The probability, \begin{align*}P\end{align*}, of an outcome, A, always falls somewhere between two extremes: (or \begin{align*}0 \%\end{align*}), which means the outcome is an impossible event and \begin{align*}1\end{align*} (or \begin{align*}100 \%\end{align*}) represents an outcome that is guaranteed to happen. These two extremes are generally not seen in real life situations. 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 between (impossible) and \begin{align*}1\end{align*} (always).

In addition, the probabilities of possible outcomes of an event must all add up to \begin{align*}1\end{align*}. This \begin{align*}1\end{align*} represents a certainty that one of the outcomes must happen. For example, tossing a coin will produce either a head or a tail. Each of these two outcomes has a probability of \begin{align*}50 \%\end{align*}, or \begin{align*}1/2\end{align*}. However, the total probabilities of the coin to land head or tail is \begin{align*}1/2 + 1/2 = 1.\end{align*}

Property 2

\begin{align*}\sum_{\text{all outcomes}} P(A) = 1\end{align*}

The sum of the probabilities of all possible outcomes must add up to \begin{align*}1\end{align*}.

Notice that tossing a coin or throwing a dice results in outcomes that are all equally probable, that is, each outcome has the same probability as the other outcome in the same sample space. Getting a head or a tail from tossing a coin produces equal probability for each outcome, \begin{align*}50 \%\end{align*}. Throwing a die also has \begin{align*}6\end{align*} possible outcomes but they all have the same probability, \begin{align*}1/6\end{align*}. We refer to this kind of probability as the classical probability. It is the simplest kind of probability. Later in this lesson, we will deal with situations where each outcome in a given sample space has different probability.

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 that the 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 to occur:

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

The following examples show you how to use this formula.


When tossing two coins, what is the probability of getting head-head \begin{align*}(HH)\end{align*}? Is the probability classical?


Since there are \begin{align*}4\end{align*} elements (outcomes) in the set of sample space: \begin{align*}\left \{HH, HT, TH, TT \right \}\end{align*}, its size then is \begin{align*}4\end{align*}. Further, there is only \begin{align*}1\end{align*} \begin{align*}HH\end{align*} outcome to occur. Using the formula above,

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

Notice that each of these \begin{align*}4\end{align*} outcomes is equally probable, namely, the probability of each is \begin{align*}1/4\end{align*}. Thus it is a classical probability. Notice also that the total probabilities of all possible outcomes add up to one: \begin{align*}1/4 + 1/4 + 1/4 + 1/4 = 1\end{align*}


What is the probability of throwing a dice and getting either \begin{align*}2, 3,\end{align*} or \begin{align*}4\end{align*}?


The sample space for a fair dice has a total of \begin{align*}6\end{align*} possible outcomes. However, the total number of outcomes for our case is \begin{align*}3\end{align*} hence,

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

So, there is a probability of \begin{align*}50 \%\end{align*} that we will get \begin{align*}2, 3,\end{align*} or \begin{align*}4\end{align*}.


Consider an experiment of tossing two coins. Assume the coins are not balanced. The design of the coins is to produce the following probabilities shown in the table:

Sample Space Probability
\begin{align*}HH\end{align*} \begin{align*}4/9\end{align*}
\begin{align*}HT\end{align*} \begin{align*}2/9\end{align*}
\begin{align*}TH\end{align*} \begin{align*}2/9\end{align*}
\begin{align*}TT\end{align*} \begin{align*}1/9\end{align*}

Figure: Probability table for flipping two weighted coins.

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


Notice that the simple events \begin{align*}HT\end{align*} and \begin{align*}TH\end{align*} 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

  1. An event is something that occurs or happens with one or more outcomes.
  2. An experiment is the process of taking a measurement or making an observation.
  3. A simple event is the simplest outcome of an experiment.
  4. The sample space is the set of all possible outcomes of an experiment, typically denoted by \begin{align*}S\end{align*}.

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?

    \begin{align*}& \text{A:} \left \{ 2\;\mathrm{on\ the\ die,\ H\ on\ the\ coin} \right \} \\ & \text{B:} \left \{ \mathrm{Even\ number\ on\ the\ die,\ T\ on\ the\ coin} \right \} \\ & \text{C:} \left \{ \mathrm{Even\ number\ on\ the\ die} \right \} \\ & \text{D:} \left \{ \mathrm{T \ on\ the\ coin} \right \}\end{align*}

  2. 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) = 2/9 \\ P(3) = P(5) = P(6) = 1/9\end{align*}
    1. Find \begin{align*}P(A)\end{align*}
    2. Find \begin{align*}P(B)\end{align*}
  3. A box contains two blue marbles and three red ones. Two marbles are drawn randomly without replacement.
    1. 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, R2,\end{align*} and \begin{align*}R3\end{align*}. List the outcomes in the sample space.
    2. Determine the probability of observing each of the following events:

    \begin{align*}& \text{A:} \left \{ 2 \;\mathrm{blue\ marbles\ are\ drawn} \right \} \\ & \text{B:} \left \{ 1 \;\mathrm{red\ and\ 1\ blue\ are\ drawn} \right \} \\ & \text{C:} \left \{ 2 \;\mathrm{red\ marbles\ are\ drawn} \right \}\end{align*}

Review Answers

  1. \begin{align*}\left \{ 1T, 1H, 2T, 2H, 3T, 3H, 4T, 4H, 5T, 5H, 6T, 6H \right \}\end{align*}
  2. A: \begin{align*}1/12\end{align*} B: \begin{align*}1/4\end{align*} C: \begin{align*}1/2\end{align*} D: \begin{align*}1/2\end{align*}
  1. \begin{align*}4/9\end{align*}
  2. \begin{align*}1/3\end{align*}
  1. \begin{align*}\big\{B1B2, B1R1, B1R2, B1R3, \\ B2B1, B2R1, B2R2, B2R3, \\ R1B1, R1B2, R1R2, R1R3,\\ R2B1, R2B2, R2R1, R2R3,\\ R3B1, R3B2, R3R1, R3R2 \big\}\end{align*}
  2. A: \begin{align*}1/10\end{align*} B: \begin{align*}3/5\end{align*} C: \begin{align*}3/10\end{align*}

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