Access this content offline
Omniguides
Skip to the reading

2.4 Events, Sample Spaces and Complements

Start with All the Possibilities

Two fair six-sided dice are rolled independently. The outcome (1, 6) differs from (6, 1): one labels the first die and the other labels the second. There are 36 possible ordered pairs. They do not overlap, and together they exhaust the possibilities. This set is the sample space, whose probability is 1.

A list of possible sums is also useful, but the sums are not equally likely. There is only one pair giving a sum of 2 and six pairs giving a sum of 7. We must count equally likely ordered pairs rather than divide by the eleven possible sums.

Worked Example: A Sum of Seven

Let EE mean “the sum is 7.” The favorable ordered pairs are shown below.

First dieSecond dieSum
167
257
347
437
527
617

For equally likely outcomes,

P(E)=outcomes in Eoutcomes in the sample space=636=16.P(E)=\frac{\text{outcomes in }E}{\text{outcomes in the sample space}}=\frac6{36}=\frac16.

About one-sixth of repeated independent rolls would have a sum of 7 in the long run. This calculation depends on the stated fair, independent-dice model. With unequal outcome probabilities, add the appropriate probabilities instead of using a simple count ratio.

The complement of EE is “the sum is not 7.” It includes every sample-space outcome outside EE. Common labels are E′E', E‾\overline E and ECE^C; they refer to the same event here. Because the event and complement partition the sample space,

P(EC)=1−P(E)=1−16=56.P(E^C)=1-P(E)=1-\frac16=\frac56.

We could count the other 30 pairs, but the complement is quicker. A complement is “not the whole event,” not merely one tempting alternative. If the event were “at least one die is 6,” its complement would be “neither die is 6,” not “both dice are 6.”

A Probability Check

Every event probability lies between 0 and 1, including the endpoints. An impossible event has probability 0; the certain sample-space event has probability 1. Our 1/61/6 and 5/65/6 are within this range and add to 1. A number such as 1.2 cannot be a probability, but a probability below 1 is normal—it does not signal a mistake. The meaningful checks are a valid sample space, correct event membership, justified probabilities and the right denominator.

Try It Yourself

The following original practice applies the ideas above to a fresh setting. Show the reasoning requested in each part. Answers and task-specific rubrics are on the separate solutions page.

Original generated practice for the Fall 2026 course.

Three Fair Coin Tosses

A fair coin is tossed independently three times. Record the tosses in order. Let E be the event “at least two heads.” This is a new complete model; all eight ordered outcomes have equal probability.

Part a

List the complete sample space. Explain why its outcomes are nonoverlapping and exhaustive, and state its total probability.

Part b

Identify every outcome in E and calculate P(E). Explain why a count ratio is justified. Write the complement using E′, an overbar, and E superscript C; describe it in words and calculate its probability in two ways.

Part c

Calculate the probability of at least one head and the probability of four heads. Explain the endpoints/range check, and show that E and its complement account for the whole sample space.

Worked answer and rubric

Glossary and deeper reading · 3 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Search lessons, terms and problems