Math Core

Simulation estimates probabilities; a handful of rules lets you compute them exactly. These rules are the grammar of probability, and nearly every later calculation in AP Statistics is built from them.

Sample spaces and events

A sample space $S$ is the list of all possible outcomes of a random process. An event is any collection of outcomes. If all outcomes are equally likely,

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

For example, rolling two fair dice has 36 equally likely outcomes. The event "sum is 7" contains $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ , so $P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$ .

Every probability model must satisfy two basic rules: each probability is between $0$ and $1$ , and the probabilities of all outcomes in the sample space add to $1$ .

The complement rule

The complement of $A$ , written $A^C$ , is the event that $A$ does not happen. Since $A$ and $A^C$ together cover the whole sample space,

$P(A^C) = 1 - P(A).$

The complement rule is a big time saver whenever "not $A$ " is easier to count than $A$ itself.

Worked example: Complement

The forecast says the probability of rain tomorrow is $0.35$ . What is the probability it does not rain?

Solution. $P(\text{no rain}) = 1 - 0.35 = 0.65$ .

Mutually exclusive events and the addition rule

Two events are mutually exclusive (or disjoint) if they can't happen at the same time: $P(A \text{ and } B) = 0$ . For mutually exclusive events,

$P(A \text{ or } B) = P(A) + P(B).$

Most events overlap, though. If you add $P(A)$ and $P(B)$ , the outcomes in both events get counted twice, so you subtract the overlap once.

General addition rule

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

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).$

"Or" in statistics means at least one of the events happens (it includes both).

A Venn diagram helps you see why. The overlap region belongs to both circles, so it is counted in $P(A)$ and again in $P(B)$ .

Events A (left) and B (right). The overlap is A and B. Outside both circles is neither.

Worked example: Streaming services

In a survey of households, $58\%$ subscribe to streaming service N, $33\%$ subscribe to service H, and $21\%$ subscribe to both. Find the probability that a randomly chosen household subscribes to (a) at least one of the two and (b) neither.

Solution.

(a) $P(N \text{ or } H) = 0.58 + 0.33 - 0.21 = 0.70$ .

(b) "Neither" is the complement of "at least one": $1 - 0.70 = 0.30$ .

Two-way tables

Data from two categorical variables are often given in a two-way table. To find a probability, divide the count of the event by the total you are choosing from.

Worked example: Probability from a two-way table

A survey of 220 students recorded grade level and phone type.

	iPhone	Android	Total
9th grade	38	22	60
10th grade	45	35	80
11th grade	52	28	80
Total	135	85	220

One student is chosen at random. Find (a) $P(\text{Android})$ , (b) $P(\text{10th grade and Android})$ , (c) $P(\text{10th grade or Android})$ .

Solution.

(a) $P(\text{Android}) = \dfrac{85}{220} = \dfrac{17}{44} \approx 0.386$ .

(b) Look at the single cell: $\dfrac{35}{220} = \dfrac{7}{44} \approx 0.159$ .

$P(\text{10th or Android}) = \dfrac{80 + 85 - 35}{220} = \dfrac{130}{220} = \dfrac{13}{22} \approx 0.591.$

You can check by adding cells directly: $45 + 35 + 22 + 28 = 130$ .

Common mistake

The most common error is forgetting to subtract the overlap in "or" problems. Only use $P(A) + P(B)$ when you are sure the events are mutually exclusive. If your "or" probability comes out bigger than $1$ , you definitely double counted.

Tip

Several related unknowns? Fill in a Venn diagram from the inside out: put $P(A \text{ and } B)$ in the overlap first, then $P(A) - P(A \text{ and } B)$ in "A only", and so on. The four regions must add to $1$ .

Practice

Practice 1

The probability that a randomly chosen adult owns a bicycle is $0.27$ . What is the probability that a randomly chosen adult does not own a bicycle?