Math Core

Every probability question starts with the same step: figure out everything that could happen. Get that list right and the rest is careful counting. Get it wrong and no formula will save you. This lesson sets up the vocabulary you'll use for the rest of the unit: sample spaces, events, complements, and the words "and" and "or."

Outcomes, sample spaces and events

When you roll a number cube, spin a spinner or draw a card, you're running an experiment. Each possible result is an outcome.

Definition

Sample space

The sample space $S$ of an experiment is the set of all possible outcomes. An event is any subset of the sample space, that is, any collection of outcomes.

For one roll of a number cube, $S = \{1, 2, 3, 4, 5, 6\}$ . The event "roll an even number" is the subset $\{2, 4, 6\}$ .

When every outcome is equally likely, probability is a ratio of counts:

P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}.

Every probability is between $0$ (impossible) and $1$ (certain), and the probabilities of all the outcomes in $S$ add up to $1$ .

Organizing a sample space

For an experiment with two stages, a table is the cleanest way to list the sample space. Put one stage down the side and the other across the top; each cell is one outcome. The number of cells is the product of the number of choices at each stage.

Worked example: Two number cubes

Roll a red cube and a blue cube. Find the probability that the sum is at least $9$ .

Each cell shows the sum (red down the side, blue across the top).

+	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

The sample space has $36$ equally likely outcomes. Count the cells with sums $9$ or more: four $9$ s, three $10$ s, two $11$ s and one $12$ , for $10$ outcomes.

P(\text{sum} \ge 9) = \frac{10}{36} = \frac{5}{18}.

Common mistake

Don't use the list of sums $\{2, 3, \dots, 12\}$ as your sample space. Those $11$ results are not equally likely: a sum of $7$ can happen $6$ ways, but a sum of $12$ only $1$ way. Always count in a sample space whose outcomes really are equally likely, like the $36$ ordered pairs.

Complements

Sometimes it's easier to count what you don't want.

Complement rule

The complement of event $A$ , written $A'$ (or $A^c$ , or "not $A$ "), is the set of outcomes in $S$ that are not in $A$ . Since $A$ and $A'$ together make up the whole sample space,

P(A') = 1 - P(A).

For two number cubes, the probability of a sum of at least $9$ is $\dfrac{5}{18}$ , so the probability of a sum of $8$ or less is $1 - \dfrac{5}{18} = \dfrac{13}{18}$ . No new counting required.

"And," "or," and overlapping events

Two events can be combined in two basic ways.

The intersection $A \cap B$ (" $A$ and $B$ ") is the set of outcomes in both events.
The union $A \cup B$ (" $A$ or $B$ ") is the set of outcomes in at least one of the events. In math, "or" includes the case where both happen.

If you add the outcomes in $A$ to the outcomes in $B$ , anything in the overlap gets counted twice. Subtract it once to fix that.

Addition rule

For any two events $A$ and $B$ ,

P(A \cup B) = P(A) + P(B) - P(A \cap B).

If $A$ and $B$ have no outcomes in common, they are mutually exclusive, $P(A \cap B) = 0$ , and the rule becomes $P(A \cup B) = P(A) + P(B)$ .

Worked example: Overlapping events with cards

A standard deck has $52$ cards: $4$ suits (hearts, diamonds, clubs, spades) of $13$ ranks each. The face cards are the jack, queen and king of each suit. You draw one card. Find the probability that it is a spade or a face card.

Spades: $13$ cards, so $P(\text{spade}) = \dfrac{13}{52}$ .
Face cards: $3 \times 4 = 12$ cards, so $P(\text{face}) = \dfrac{12}{52}$ .
Both: the jack, queen and king of spades, so $P(\text{spade and face}) = \dfrac{3}{52}$ .

P(\text{spade or face}) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}.

If you had just added $13 + 12 = 25$ , you would have counted the three spade face cards twice.

Geometric probability

In geometry, outcomes are often points in a region rather than items on a list. If a point is chosen at random in a region, each small piece of the region is equally likely to contain it, so you compare sizes (lengths or areas) instead of counts.

P(\text{point lands in region } R) = \frac{\text{area of } R}{\text{area of the whole region}}.

Worked example: A dartboard

A square board is $12$ inches on each side. A circle of radius $4$ inches is painted in the middle. A dart hits the board at a random point. What is the probability that it lands inside the circle?

P = \frac{\pi (4)^2}{12^2} = \frac{16\pi}{144} = \frac{\pi}{9} \approx 0.349.

The probability that the dart misses the circle is the complement, $1 - \dfrac{\pi}{9} \approx 0.651$ .

Tip

Before you count favorable outcomes, check the size of your sample space. For a two-stage experiment it should equal (choices for stage 1) $\times$ (choices for stage 2). If your table for a coin and a spinner with $5$ sections has $9$ cells, one is missing.

Practice

Practice 1

You flip a coin and spin a spinner with $5$ equal sections numbered $1$ to $5$ . How many outcomes are in the sample space?