Any process that terminates in one or more outcomes (results) is an experiment. For example, if we ask people what time of the day they typically eat their first meal or if we pull balls from a bag of colored balls, we have performed an experiment. The outcomes of the first experiment are the times of day given as responses to our question. The outcomes of the second experiment are the colors of the balls drawn from the bag.

The set of all possible outcomes of a given experiment is called the sample space of the experiment. Suppose the experiment is tossing a coin. Then we could denote the outcome “heads” by H; denote the outcome “tails” by T; and write the sample space, S, in set notation as

A single die is a cube, each face of which contains one, two, three, four, five, or six dots and no two faces contain the same number of dots. A roll of the die is an experiment, the outcome of which is the number of dots showing on the upper face, and the sample space S can be written in set notation as

An event is any set of possible outcomes; that is, an event is any subset of a sample space. For the die-rolling experiment, one possible event is “The number showing is even.” In set notation, we write this event as $E=\{2,4,6\}.$ The event “The number showing is a 6” is written in set notation as $E=\left\{6\right\}.$

Outcomes of an experiment are said to be equally likely if no outcome should result more often than any other outcome when the experiment is run repeatedly. In both the coin-tossing and die-rolling experiments, all of the outcomes are equally likely.

Because E is a subset of S, $0\le n(E)\le n(S).$ Dividing by n(S) yields $0\le {\displaystyle \frac{n(E)}{n(S)}}\le 1.$

So the probability P(E) of an event E is a number between 0 and 1 inclusive.

Notice in Example 1d that $E_4=S;$ so the event $E_4$ is certain to occur on every roll of the die. Every certain event has probability 1. On the other hand, in Example 1e, the event $E_5=\varnothing ;$ so $E_5$ never occurs, and $P(E_5)=0.$ An impossible event always has probability 0.

Equally likely outcomes occur in most games of chance that involve tossing coins, rolling dice, or drawing cards. Lotteries and games involving spinning wheels also are assumed to generate equally likely outcomes.

Example 3 Rolling a Pair of Dice

What is the probability of getting a sum of 8 when a pair of dice is rolled?

Solution

There are six equally likely outcomes for each die; so by the Fundamental Counting Principle, there are $6\cdot 6$, or 36, equally likely outcomes when a pair of dice is rolled. It is useful to think of having two dice of different colors—say, red and green—in distinguishing the 36 outcomes. In this way, 2 on the red die and 6 on the green die is easily seen to be a different physical outcome from 6 on the red die and 2 on the green die, even though the same sum results.

We can represent the 36 outcomes as (red die, green die) pairs:

The highlighted ordered pairs in the sample space are the outcomes that have a sum of 8. In set notation, the event “The sum of the numbers on the dice is 8” is

$$E=\{(6,2),(5,3),(4,4),(3,5),(2,6)\}.$$

Because $n(E)=5$ and $n(S)=36,$

$$P(E)={\displaystyle {\displaystyle \frac{n(E)}{n(S)}}}={\displaystyle {\displaystyle \frac{5}{36}}}.$$

So the probability of getting a sum of 8 is ${\displaystyle \frac{5}{36}}.$

Practice Problem 3

1. What is the probability of getting a sum of 7 when a pair of dice is rolled?

Consider the experiment of rolling one die. The sample space is $S=\{1,2,3,4,5,6\},$ and $n(S)=6.$

Let E be the event “The number rolled is greater than 4” and let F be the event “The number rolled is even.” Then

The event “The number rolled is greater than 4 and even” is $E\cap F.$

Because $n(E)=2,n(F)=3,n(E\cap F)=1,$ and $n(S)=6,$ we have

The event “The number rolled is either greater than 4 or even” is $E\cup F.$

Note that 6 is in both E and F, so it is counted once when n(E) is counted and a second time when n(F) is counted. However, 6 is counted only once when $n(E\cup F)$ is counted. We have $n(E\cup F)=n(E)+n(F)-n(E\cap F)$. That is, because 6 was counted twice in adding $n(E)+n(F),$ we must subtract $n(E\cap F)$ to get an accurate count for $E\cup F.$

This example illustrates the Additive Rule.

Two events are mutually exclusive if it is impossible for both to occur simultaneously. That is, E and F are mutually exclusive events if $E\cap F$ contains no outcomes. In this case, $P(E\cap F)=0$ and the Additive Rule is somewhat simplified.

A standard deck of 52 playing cards uses four different designs called suits:

Each suit has 13 cards: an ace, a king, a queen, and a jack, and cards numbered 2 through 10. The diamond and heart symbols are red, and the club and spade symbols are black. For the purposes of this section, we assume that all decks of cards are standard and well shuffled, so that all cards drawn or dealt from a deck are equally likely.

Example 5 Drawing a Card from a Deck

A card is drawn from a deck. What is the probability that the card is

1. A king?

2. A spade?

3. A king or a spade?

4. A heart or a spade?

Solution

The sample space S is the 52-card deck; so $n(S)=52.$

1. There are four kings, one in each suit; so

   $$P(\text{king})={\displaystyle {\displaystyle \frac{4}{52}}}={\displaystyle {\displaystyle \frac{1}{13}}}.$$

2. There are 13 spades in the deck; so

   $$P(\text{spade})={\displaystyle {\displaystyle \frac{13}{52}}}={\displaystyle {\displaystyle \frac{1}{4}}}.$$

3. By the Additive Rule, we have

   $$P(\text{king or spade})=P(\text{king})+P(\text{spade})-P(\text{king and spade}).$$

   Because there is only one card that is both a king and a spade,

   $P(\text{king and spade})={\displaystyle \frac{1}{52}}.$ From parts a and b in this example,

   $$\begin{array}{rcl}P(\text{king or spade})& =& {\displaystyle {\displaystyle \frac{4}{52}}}+{\displaystyle {\displaystyle \frac{13}{52}}}-{\displaystyle {\displaystyle \frac{1}{52}}}\\ & =& {\displaystyle {\displaystyle \frac{16}{52}}}={\displaystyle {\displaystyle \frac{4}{13}}}\end{array}$$

4. Because no card is both a heart and a spade, the events “Draw a heart” and “Draw a spade” are mutually exclusive. Thus,

   $$\begin{array}{rcl}P(\text{heart or spade})& =& P(\text{heart})+P(\text{spade})\\ & =& {\displaystyle {\displaystyle \frac{13}{52}}}+{\displaystyle {\displaystyle \frac{13}{52}}}={\displaystyle {\displaystyle \frac{1}{2}}}\end{array}$$

Practice Problem 5

1. What is the probability that a card pulled from a deck is a jack or a king?

The complement of an event E is the set of all outcomes in the sample space that are not in E. The complement of E is denoted by $E^{\prime }$. Because every outcome in the sample space is in E or in $E^{\prime }$, the event $E\cup E^{\prime }$ is a certain event and $P(E\cup E^{\prime })=1.$ Also, E and $E^{\prime }$ are mutually exclusive events $(E\cap E^{\prime }=\varnothing ,$ and $P(E\cap E^{\prime })=0)$, so $P(E\cup E^{\prime })=P(E)+P(E^{\prime }).$ Thus, $P(E)+P(E^{\prime })=1$ and $P(E^{\prime })=1-P(E).$

The probabilities we have seen up to this point have been determined by some form of reasoning, not data. Data support the decision to assign the value ${\displaystyle \frac{1}{2}}$ as the probability of a head resulting from one toss of a coin, but the assignment is not based on data. This is also true for the probabilities associated with drawing a card, rolling a pair of dice, drawing a ball from an urn, and so on. Probability assigned this way is called theoretical probability.

In contrast, we can assign probability values based on observation and data. If an experiment is performed n times (where n is large) and a particular outcome occurs k times, then the number ${\displaystyle \frac{k}{n}}$ is the value for the probability of that outcome. Probability determined this way is called experimental probability. A statement such as “The probability that a 10-year-old child will live to be 95 years old is ${\displaystyle \frac{3}{\mathrm{100,000}}}$ ” refers to experimental probability.