Sample Spaces and Simple Probability | Geometry | LevelKite

Before you can say how likely something is, you need to know everything that could happen. Three terms describe this.

An experiment is an action whose result you cannot be sure of ahead of time, such as rolling a standard die (a cube with faces numbered $1$ to $6$ ).
An outcome is one possible result of an experiment.
The sample space, written $S$ , is the set of all possible outcomes.

A complete sample space lists every outcome exactly once. The table shows three examples, each written as a set inside braces.

Sample spaces of three simple experiments

Experiment	Sample space	Number of outcomes
Flip one coin ( $H$ for heads, $T$ for tails)	$S = \{H, T\}$	$2$
Roll one standard die	$S = \{1, 2, 3, 4, 5, 6\}$	$6$
Choose one letter from the word MATH	$S = \{M, A, T, H\}$	$4$

Worked example

Listing a sample space

You flip a coin twice and record each result in order. List the sample space and count its outcomes.

Decide what one outcome looks like. An outcome is one possible result of the whole experiment, and this experiment has two flips. So one outcome records both flips in order. For example, HT means heads first, then tails.
List the outcomes that start with heads. The first flip is H, and the second flip can be H or T, since those are the two results of a flip.
$HH, \; HT$
List the outcomes that start with tails in the same way. Working through the first flip in order keeps you from missing or repeating an outcome.
$TH, \; TT$
Collect all the outcomes into one set, because the definition of sample space calls for all possible outcomes. HT and TH are different outcomes because their first flips differ.
$S = \{HH, HT, TH, TT\}$
Count the outcomes in the set. Each one appears exactly once, so the count is correct.
$\text{number of outcomes} = 4$

Answer

The sample space is $S = \{HH, HT, TH, TT\}$ , and it has $4$ outcomes.

Once you have the sample space, you can ask about part of it. An event is any set of outcomes from the sample space. That makes it a subset of $S$ , meaning a set whose members all belong to $S$ . The event happens when the result of the experiment is one of its outcomes.

For one roll of a die, the event "rolling an even number" is $E = \{2, 4, 6\}$ . In this lesson, a simple event is an event from one experiment, with one outcome or several. Some books save that name for an event with exactly one outcome, so this lesson shows both kinds.

Probability is a number that measures how likely an event is. Its meaning is clearest when outcomes are equally likely, which means each outcome has the same chance of happening. Then the probability of an event is the fraction of the sample space that the event covers.

Use this idea only when the problem gives you a reason to. A coin or die is fair when its outcomes are equally likely, and choosing at random means every item has the same chance of being chosen. Those stated words are your reason. Never assume equal chances just because a list looks tidy.

Key concept

Probability with equally likely outcomes

Suppose a sample space $S$ has a finite number of equally likely outcomes and at least one outcome. The probability of an event $E$ is the number of outcomes in $E$ divided by the number of outcomes in $S$ . You write these counts as $n(E)$ and $n(S)$ . The denominator $n(S)$ must be nonzero, because division by zero is undefined.

P(E) = \frac{n(E)}{n(S)}, \quad n(S) \neq 0

This definition gives you a check for every probability you compute. All the outcomes in $E$ come from $S$ , so $0 \le n(E) \le n(S)$ . Because $S$ has at least one outcome, $n(S)$ is positive, and so it is nonzero. Divide all three parts by $n(S)$ . Dividing by a positive number keeps the inequalities pointing the same way, so $\frac{0}{n(S)} \le \frac{n(E)}{n(S)} \le \frac{n(S)}{n(S)}$ . This simplifies to $0 \le P(E) \le 1$ .

So a probability is never negative and never greater than $1$ . An event with no outcomes has probability $0$ , and an event that contains every outcome in $S$ has probability $1$ .

Worked example

Finding the probability of a simple event

You roll a fair standard die once. Find the probability of each event: (a) rolling a $5$ , and (b) rolling a number greater than $4$ .

Write the sample space, using its definition: the set of all possible outcomes. The word fair is given, so by the definition of a fair die the 6 outcomes are equally likely. That lets you use the definition of probability.
$S = \{1, 2, 3, 4, 5, 6\}, \quad n(S) = 6$
For part (a), write the event as a subset of S. Only one outcome is a 5.
$A = \{5\}, \quad n(A) = 1$
Apply the definition of probability. The denominator, 6, is nonzero, so the fraction is defined.
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{6}$
For part (b), write the event as a subset of S. The numbers 5 and 6 are greater than 4. The number 4 is not, because 4 is not greater than itself.
$B = \{5, 6\}, \quad n(B) = 2$
Apply the definition of probability again. Simplify by dividing the top and bottom by their common factor, 2, which is nonzero.
$P(B) = \frac{n(B)}{n(S)} = \frac{2}{6} = \frac{2 \div 2}{6 \div 2} = \frac{1}{3}$
Check both answers against the fact that every probability is from 0 to 1. Both fractions pass.
$0 \le \frac{1}{6} \le 1 \quad \text{and} \quad 0 \le \frac{1}{3} \le 1$

Answer

The probability of rolling a $5$ is $\frac{1}{6}$ , and the probability of rolling a number greater than $4$ is $\frac{1}{3}$ .

Sometimes you care about an event not happening. The complement of an event $E$ , written $E'$ and read "not $E$ ," is the set of all outcomes in $S$ that are not in $E$ . For one roll of a die, if $E = \{5, 6\}$ , then $E' = \{1, 2, 3, 4\}$ .

Every outcome in $S$ lies in exactly one of these two sets. Together they cover the whole sample space with no overlap. So when $n(S) \neq 0$ , the fractions of the sample space they cover add up to the whole, which is $1$ .

This idea is the complement rule. When the outcomes of $S$ are equally likely and $n(S) \neq 0$ , the probability that $E$ does not happen is $1$ minus the probability that $E$ happens. The rule is below, followed by its proof.

P(E') = 1 - P(E), \quad n(S) \neq 0

Proof of the complement rule

Statement	Reason
$S$ has equally likely outcomes, $n(S) \neq 0$ , and $E$ is an event in $S$	Given
$n(E) + n(E') = n(S)$	Definition of complement: each outcome of $S$ is in exactly one of $E$ and $E'$
$\frac{n(E) + n(E')}{n(S)} = \frac{n(S)}{n(S)}$	Division Property of Equality, allowed because $n(S) \neq 0$
$\frac{n(E)}{n(S)} + \frac{n(E')}{n(S)} = 1$	Adding fractions with a common denominator, and $\frac{n(S)}{n(S)} = 1$ because $n(S) \neq 0$
$P(E) + P(E') = 1$	Definition of probability, applied to $E$ and to $E'$
$P(E') = 1 - P(E)$	Subtraction Property of Equality: subtract $P(E)$ from both sides

Worked example

Using the complement rule

You choose one card at random from $10$ cards numbered $1$ through $10$ . Find the probability that the number on the card is not a multiple of $3$ .

Write the sample space, the set of all possible outcomes. The phrase at random is given, so by its definition the 10 outcomes are equally likely.
$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, \quad n(S) = 10$
Let E be the event that the number is a multiple of 3. You want its complement, E'. E has the shorter list, so count E first.
$E = \{3, 6, 9\}, \quad n(E) = 3$
Apply the definition of probability. The denominator, 10, is nonzero.
$P(E) = \frac{n(E)}{n(S)} = \frac{3}{10}$
Apply the complement rule. Its conditions hold because the outcomes are equally likely and n(S) is nonzero. Write 1 as 10/10, which is valid because 10 is nonzero, then subtract the numerators.
$P(E') = 1 - \frac{3}{10} = \frac{10}{10} - \frac{3}{10} = \frac{7}{10}$
Check by listing E' directly from the definition of complement. The count agrees, and 7/10 is between 0 and 1.
$E' = \{1, 2, 4, 5, 7, 8, 10\}, \quad \frac{n(E')}{n(S)} = \frac{7}{10}$

Answer

The probability that the number is not a multiple of $3$ is $\frac{7}{10}$ .

A word problem uses the same definitions in everyday language. Start by separating what is given from what you must find. Then answer three questions:

What is one outcome? Pick outcomes that are equally likely. When a person or object is chosen at random, each person or object is one outcome. A category, such as a grade level, is not.
How many outcomes are there? This count is $n(S)$ , and it must be nonzero for the fraction to be defined.
Which is easier to count, the event or its complement? Count that one first. If it is the complement, use the complement rule.

Worked example

A word problem: choosing a speaker

A robotics club has $24$ members: $9$ ninth graders, $7$ tenth graders, $5$ eleventh graders and $3$ twelfth graders. Each member's name goes into a bag once, and the club draws one name at random to choose a speaker. What is the probability that the speaker is not a ninth grader?

Separate what is given from what you must find. Given: 24 names, each in the bag once, and a random draw. To find: the probability that the speaker is not a ninth grader.
Identify the outcomes. By the definition of at random, each member is one equally likely outcome, so S has 24 outcomes, not 4 grades. Each member's probability of being chosen is 1/24, so the draw is fair to everyone.
$n(S) = 9 + 7 + 5 + 3 = 24$
Let E be the event that the speaker is a ninth grader. E has 9 outcomes and n(S) is nonzero, so use the definition of probability.
$P(E) = \frac{n(E)}{n(S)} = \frac{9}{24}$
The question asks about E', so apply the complement rule. Its conditions hold because the outcomes are equally likely and n(S) is nonzero. Write 1 as 24/24, which is valid because 24 is nonzero, then subtract the numerators.
$P(E') = 1 - \frac{9}{24} = \frac{24}{24} - \frac{9}{24} = \frac{15}{24}$
Simplify by dividing the top and bottom by their common factor, 3, which is nonzero.
$\frac{15}{24} = \frac{15 \div 3}{24 \div 3} = \frac{5}{8}$
Check by counting E' directly from the definition of complement: the other three grades. The count agrees, and 5/8 is between 0 and 1.
$\frac{n(E')}{n(S)} = \frac{7 + 5 + 3}{24} = \frac{15}{24} = \frac{5}{8}$

Answer

The probability that the speaker is not a ninth grader is $\frac{5}{8}$ .

Probability starts with a complete sample space. When its outcomes are equally likely and $n(S) \neq 0$ , the fraction $P(E) = \frac{n(E)}{n(S)}$ tells you how much of the sample space an event covers, and $P(E') = 1 - P(E)$ tells you how much is left.