The Multiplication Principle for Counting Outcomes

Counting Complete Outcomes One Stage at a Time

In Sample Spaces and Outcomes, you learned that a sample space is the set of all possible complete outcomes of a chance process. Sometimes a process has many stages, and listing every complete outcome would be tedious. The multiplication principle gives a faster way to count them. It also lets you count the total outcomes that can serve as a probability denominator when those outcomes are equally likely.

Definition: If a process has stages with \(a_1, a_2, \ldots, a_k\) choices at each stage, and the number of choices at each stage is the same for every way of completing the earlier stages, then the total number of complete outcomes is the product \(a_1 \times a_2 \times \cdots \times a_k\).

\text{Number of complete outcomes}=a_1 \times a_2 \times \cdots \times a_k

The stages could be choosing a top, then pants, then shoes; or entering the first character, then the second, then the third in a code. Each complete outcome records one choice at every stage. Multiplication counts all the possible ways to combine those choices.

Pay attention to what counts as a distinct outcome. In an outfit, a particular top with particular pants and particular shoes is one complete outfit. In a password, the order of characters matters: AB and BA are different passwords. Also check the rules about repeating choices. If a character can be used again, it may still be available at the next position; if it cannot, the number of available choices may decrease.

The multiplication principle counts outcomes; by itself, it does not say they are equally likely. The counting rule from Equally Likely Outcomes and Counting Probability applies only when each outcome in the sample space has the same probability. For example, a randomly generated code with each digit chosen uniformly and independently has equally likely codes. A count of outfits does not imply equal likelihood if some outfits are more likely to be selected than others.

Set Up the Stages Before Multiplying

A useful first step is to describe the stages in order and record how many choices are available at each one. Then check that those counts apply no matter which choices were made earlier. If an earlier choice changes the number of later choices, a single product may not work without more careful counting. In this tutorial, focus on processes where the number of choices at each stage is fixed.

Define one complete outcome.
Say exactly what information makes two results different, such as the selected clothing items or the ordered characters in a code.

Identify the stages.
List the choices made to produce one complete result, in a sensible order.

Count choices at each stage.
Check whether choices may repeat and whether earlier choices affect the number available later.

Multiply and interpret.
Multiply the stage counts to get the number of complete outcomes. If using that count for probability, also verify that the outcomes are equally likely.

When finding the probability of an event, use the same idea to count outcomes in the event. If all complete outcomes are equally likely, divide the number of favorable outcomes by the total number of outcomes. The favorable count must refer to complete outcomes too—not just to choices at one stage.

Formula: When the complete outcomes are equally likely, the probability of an event is the number of favorable complete outcomes divided by the total number of complete outcomes.

P(\text{event}) = \frac{\text{number of favorable outcomes}} {\text{total number of outcomes}}

Worked Examples: Outfits and Codes

The first examples show how to count every possible combination, then count only the combinations that meet an event description. In each probability calculation, the model specifies how outcomes are selected so that equal likelihood is justified.

Worked Example: Counting Outfits and Finding a Probability

A student is choosing an outfit from 4 tops, 3 pairs of pants, and 2 pairs of shoes. Each item is selected uniformly at random from its category, independently of the other selections. How many outfits are possible? What is the probability that the outfit includes one of the 2 blue tops?

Define the outcome and stages: One outcome is a complete outfit: one top, one pair of pants, and one pair of shoes. These are three stages, with 4, 3, and 2 choices, respectively.

Count all outfits:

4 \times 3 \times 2=24

There are 24 possible complete outfits. Because each item is selected uniformly from its category and the selections are independent, each of the 24 combinations has probability \(1/24\). Thus, the outcomes are equally likely.

Count favorable outfits: There are 2 choices of a blue top, 3 choices of pants, and 2 choices of shoes. The number of outfits with a blue top is:

2 \times 3 \times 2=12

Find the probability:

P(\text{blue top}) = \frac{12}{24} = \frac{1}{2} = 0.50

Conclude: There are 24 possible outfits, and the probability of selecting an outfit with a blue top is \(0.50\), or 50%. This probability uses the count of complete outfits as its denominator because the model makes all 24 outfits equally likely.

Worked Example: A Four-Digit PIN

An invented lock generates a four-digit PIN. Each position can contain any digit from 0 through 9, digits may repeat, and a PIN may begin with 0. If the lock chooses each digit uniformly and independently, what is the probability that the first digit is 7 and the last digit is even?

State: The sample space consists of all ordered four-digit strings, including strings such as 0042. The event is that the first digit is 7 and the fourth digit is one of 0, 2, 4, 6, or 8.

Plan: Count all valid PINs by multiplying the choices for the four positions. Then count favorable PINs by applying the event restrictions at the first and fourth positions while leaving the two middle positions unrestricted. Since digits are chosen uniformly and independently, all PINs are equally likely, so the favorable count divided by the total count gives the probability.

Do—count the sample space: Each of the four positions has 10 possible digits. Repetition and a leading zero are allowed.

10 \times 10 \times 10 \times 10=10{,}000

Do—count favorable PINs: The first position has 1 choice (7), each middle position has 10 choices, and the last position has 5 even-digit choices.

1 \times 10 \times 10 \times 5=500

Therefore:

P(\text{first digit is 7 and last digit is even}) = \frac{500}{10{,}000} = \frac{1}{20} = 0.05

Conclude: The probability that the generated PIN starts with 7 and ends with an even digit is \(0.05\), or 5%. The denominator is 10,000 because that is the number of equally likely PINs allowed by the stated rules.

When Choices Cannot Be Repeated

If repetition is not allowed, count the available choices again at each stage. The count may decrease after each selection. Do not keep using the first stage’s number of choices if a previously selected item is no longer available.

Worked Example: A Password with No Repeated Letters

An invented password generator makes a four-character password from three positions for letters followed by one position for a symbol. The first three positions use letters from a set of 8 letters, and no letter can be used more than once. The last position uses one of 4 symbols. The generator chooses uniformly from all valid passwords. What is the probability that the password begins with a particular letter and ends with one of 2 specified symbols?

Define the stages: The password has three ordered letter positions and a final symbol position. Order matters, so using the same three letters in a different order produces a different password. Repetition is prohibited for the letter positions.

Count all valid passwords: There are 8 choices for the first letter. After choosing it, 7 letters remain for the second position; then 6 remain for the third. The final symbol has 4 choices.

8 \times 7 \times 6 \times 4 = 336 \times 4 = 1{,}344

Count favorable passwords: The first letter is fixed, so there is 1 choice for that position. There are 7 remaining choices for the second letter and 6 for the third. The final symbol can be either of the 2 specified symbols.

1 \times 7 \times 6 \times 2 = 84

Find and interpret the probability:

P(\text{specified first letter and one of the two symbols}) = \frac{84}{1{,}344} = \frac{1}{16} = 0.0625

Conclude: The probability is \(0.0625\), or 6.25%, under this generator. The reduction in letter choices matters: using \(8 \times 8 \times 8 \times 4\) would count passwords with repeated letters, which are not allowed.

This example also illustrates why order matters. A password records which character is in each position. Counting only sets of three letters would ignore their positions and would not count the password outcomes defined in the question.

Check Whether a Counting Result Is a Valid Denominator

A multiplication result can be correct as a count and still be unsuitable as the denominator in a favorable-over-total probability calculation. That calculation relies on equally likely outcomes. Consider a process that selects one of 3 tops and one of 2 pairs of pants, but chooses the first top with probability one-half and each other top with probability one-quarter. There are still \(3 \times 2=6\) outfits, but those outfits are not all equally likely. Counting favorable outfits and dividing by 6 would not generally give the probability. Instead, you would need the probabilities of the relevant outcomes.

A good solution therefore separates two questions: How many complete outcomes are possible? And are those outcomes equally likely under the chance model? State the multiplication count clearly, then justify its use as a denominator rather than assuming that every list of choices creates equal probabilities.

Common Mistakes and AP Exam Tips

Adding stage counts instead of multiplying. If you choose one item at each of several stages, multiply the numbers of choices to count complete combinations. Addition is used for separate alternatives, not for combining one choice from every stage.
Forgetting that positions matter. In a code or password, changing the order usually creates a different outcome. Count choices for each position.
Ignoring repetition rules. If repetition is allowed, the same number of choices may remain at later positions. If repetition is prohibited, update the number of available choices after each selection.
Assuming a leading zero is forbidden. A PIN is often a string of digits, not a number. Follow the stated rules; if zero is permitted in the first position, include it among the choices.
Using a count as a probability denominator without checking equal likelihood. The multiplication principle gives the size of the sample space, not the probabilities of its outcomes. A favorable-over-total ratio is justified only when the complete outcomes are equally likely.
Counting partial results as favorable outcomes. If an event concerns complete outfits or passwords, count the complete combinations that satisfy it, with every stage represented.

For a full-credit response, identify what one outcome represents, show the number of choices at each stage, and multiply to count complete outcomes. For a probability, show the favorable count as well as the total count, and explain why the outcomes are equally likely under the stated model.

Key takeaway: Count a multi-stage outcome by multiplying the choices available at each stage, adjusting for the rules on order and repetition. Use that total as a probability denominator only when the complete outcomes are equally likely.

Check Your Understanding

For each question, describe the stages before calculating. For probability questions, check whether the model makes the complete outcomes equally likely.

A student chooses one of 5 shirts, one of 2 skirts, and one of 3 pairs of shoes. How many outfits are possible?
A three-digit code uses digits 0 through 9. Repetition is allowed, and the code may begin with 0. How many codes are possible?
A four-character password uses letters in all four positions, selected from 6 allowed letters with no repetition. How many passwords are possible?
A randomly generated two-digit code uses digits 0 through 9 independently and uniformly, with repetition allowed. What is the probability that both digits are 4?
A selection process has 4 possible complete outcomes, but one outcome is twice as likely as each of the other three. Is the probability of an event necessarily its number of favorable outcomes divided by 4? Explain.