Math Core

Many chance situations have more than one part: flip a coin and roll a number cube, or spin two spinners. An event built from two or more parts is a compound event. The key is to list the whole sample space in an organized way, so you don't miss or double-count any outcome.

Compound events and their sample spaces

Definition

Compound event

A compound event is an event that depends on two or more actions, such as rolling two number cubes. Each outcome of the compound experiment is a combination, like (heads, $4$ ).

Once you have the full sample space, finding a probability works just like before.

Probability of a compound event

List every outcome of the compound experiment (using an organized list, a table or a tree diagram). If the outcomes are equally likely,

P(\text{event}) = \frac{\text{number of outcomes in the event}}{\text{total number of outcomes}}.

Using a table

A table works well when there are exactly two parts.

Worked example: A coin and a number cube

You flip a coin and roll a number cube. What is the probability of getting heads and an even number?

Make a table with the coin down the side and the number cube across the top.

	$1$	$2$	$3$	$4$	$5$	$6$
H	H1	H2	H3	H4	H5	H6
T	T1	T2	T3	T4	T5	T6

There are $2 \times 6 = 12$ outcomes. Heads with an even number: H2, H4, H6. That's $3$ outcomes, so

P(\text{heads and even}) = \frac{3}{12} = \frac{1}{4}.

Notice that the number of outcomes is the number of choices for the first part times the number of choices for the second part. That's a quick way to know how big your sample space should be.

Worked example: Two number cubes

You roll two number cubes, one red and one blue, and add the numbers. What is the probability that the sum is $8$ ?

Each cell shows the sum of the red cube (side) and the blue cube (top).

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

There are $6 \times 6 = 36$ outcomes. A sum of $8$ appears $5$ times: $(2, 6)$ , $(3, 5)$ , $(4, 4)$ , $(5, 3)$ , $(6, 2)$ . So $P(\text{sum} = 8) = \dfrac{5}{36}$ .

Common mistake

Red $2$ with blue $6$ and red $6$ with blue $2$ are different outcomes. If you count them as one, you'll get the wrong total. For the same reason, the sums $2$ through $12$ are not equally likely: a sum of $7$ happens $6$ ways, but a sum of $2$ happens only $1$ way.

Using a tree diagram

When there are three or more parts, a tree diagram keeps things organized. Each branch splits into every possible next result, and each path from start to finish is one outcome.

Worked example: Three coin flips

You flip a coin three times. What is the probability of getting exactly two heads?

Build the tree one flip at a time.

H
- H: HHH, HHT
- T: HTH, HTT
T
- H: THH, THT
- T: TTH, TTT

There are $2 \times 2 \times 2 = 8$ outcomes. Exactly two heads: HHT, HTH, THH. So

P(\text{exactly two heads}) = \frac{3}{8}.

Tip

Before you count favorable outcomes, check the size of your sample space by multiplying the number of choices for each part. If you listed $7$ outcomes for three coin flips, you know one is missing.