Independent Events and Unions of Events (AP Statistics 2.7)

A gym has 400 members, and 25% of them take group fitness classes. Among monthly members, the share taking classes is also 25%, and among annual members it's 25% again. Learning someone's membership type tells you nothing new about whether they take classes. That is what it means for two events to be independent, and it makes "both" and "either" probabilities easy to find.

Independent events

Events $A$ and $B$ are independent when knowing whether one of them happened does not change the probability of the other. In symbols, using conditional probability from topic 2.6:

Here are the gym's data:

	Classes	No classes	Total
Monthly	60	180	240
Annual	40	120	160
Total	100	300	400

For a randomly chosen member, $P(\text{classes}) = \frac{100}{400} = 0.25$ and $P(\text{classes} \mid \text{monthly}) = \frac{60}{240} = 0.25$ . They're equal, so "takes classes" and "monthly member" are independent. To show independence from a table, compute one conditional probability and the matching unconditional one, and compare them.

Independence usually comes from how the process works rather than from a table. Two dice rolled together, a coin flipped twice, or two people drawn from a hat when the first name is put back before the second draw: the first result can't affect the second, so the events are independent. Drawing without putting back is different (topic 2.6), because the first result changes what's left.

With data, the comparison has to come out equal. If $P(A \mid B)$ is 0.36 and $P(A)$ is 0.30, the events are dependent for these individuals, even though the numbers are near each other. Whether a difference like that is large enough to be evidence about a whole population is a question for inference in Unit 3.

The multiplication rule for independent events

The general multiplication rule says $P(A \cap B) = P(A) \cdot P(B \mid A)$ . If the events are independent, $P(B \mid A)$ is just $P(B)$ , so:

P(A \cap B) = P(A) \cdot P(B) \quad \text{when } A \text{ and } B \text{ are independent.}

This works in both directions. Given independence, multiply to get the joint probability. Given all three numbers, check whether $P(A \cap B)$ equals $P(A) \cdot P(B)$ : if it does, the events are independent, and if not, they are dependent. For the gym, $P(\text{monthly}) \cdot P(\text{classes}) = 0.6 \cdot 0.25 = 0.15$ , and the table gives $P(\text{monthly} \cap \text{classes}) = \frac{60}{400} = 0.15$ , which agrees with the first check.

The rule extends to more than two independent events: multiply all their probabilities. That gives the standard way to find "at least one": its complement is "none," which is a product. If each of three independent traffic lights on your route is red with probability 0.4, then

P(\text{at least one red}) = 1 - P(\text{no reds}) = 1 - 0.6^3 = 1 - 0.216 = 0.784.

Unions: one event or the other

The union of $A$ and $B$ , written $A \cup B$ , is the event that $A$ happens, or $B$ happens, or both. In probability, "or" always includes "both" unless a question says otherwise.

To count the outcomes in $A \cup B$ , you could add the outcomes in $A$ and the outcomes in $B$ . That counts every outcome in both events twice, once in each, so you subtract the overlap once:

For the gym, $P(\text{monthly} \cup \text{classes}) = 0.6 + 0.25 - 0.15 = 0.70$ . You can confirm it from the table: the only members in neither event are the 120 annual members who skip classes, so the union is $1 - \frac{120}{400} = 0.70$ .

The complement of " $A$ or $B$ " is "neither $A$ nor $B$ ," so $P(A \cup B) = 1 - P(\text{neither})$ . When one cell of a table holds "neither," it's often the fastest route.

Independent is not the same as mutually exclusive

The two words sound like they describe events that "have nothing to do with each other," and students swap them. They mean nearly opposite things:

Mutually exclusive (topic 2.5): the events can't happen together, $P(A \cap B) = 0$ .
Independent: one event happening doesn't change the other's probability, $P(A \cap B) = P(A) \cdot P(B)$ .

If $P(A)$ and $P(B)$ are both greater than 0, then $P(A) \cdot P(B) > 0$ , so mutually exclusive events (joint probability 0) can never be independent. Learning that $A$ happened drops $B$ 's probability all the way to 0, which is as strong a dependence as there is. In the gym table, "monthly" and "annual" are mutually exclusive, and $P(\text{annual} \mid \text{monthly}) = 0$ while $P(\text{annual}) = 0.4$ .

Worked examples

Worked exampleExample 1: testing a table for independence

A survey asked 500 adults in a city whether they own a smart speaker.

	Owns one	Doesn't	Total
Under 40	90	160	250
40 or older	60	190	250
Total	150	350	500

One adult from the survey is chosen at random. (a) Are "under 40" and "owns a smart speaker" independent? Justify. (b) Find the probability that the adult is under 40 or owns a smart speaker.

Show solution

(a) $P(\text{owns}) = \frac{150}{500} = 0.30$ and $P(\text{owns} \mid \text{under 40}) = \frac{90}{250} = 0.36$ . Since $0.36 \neq 0.30$ , knowing the adult is under 40 changes the probability of owning a smart speaker, so the events are not independent.

(b) By the addition rule, with $P(\text{under 40}) = \frac{250}{500} = 0.5$ and $P(\text{under 40} \cap \text{owns}) = \frac{90}{500} = 0.18$ :

P(\text{under 40} \cup \text{owns}) = 0.5 + 0.30 - 0.18 = 0.62.

Check with the complement: the 190 adults 40 or older who don't own one are in neither event, and $1 - \frac{190}{500} = 0.62$ .

Worked exampleExample 2: two smoke detectors

A house has two smoke detectors that work independently. In a fire, the hall detector sounds with probability 0.90 and the kitchen detector with probability 0.85. Find the probability that in a fire (a) both sound, (b) at least one sounds, and (c) exactly one sounds.

Show solution

(a) The detectors are independent, so multiply: $P(H \cap K) = 0.90 \cdot 0.85 = 0.765$ .

(b) "At least one" is the union. Addition rule: $P(H \cup K) = 0.90 + 0.85 - 0.765 = 0.985$ . The complement route agrees: neither sounds with probability $0.10 \cdot 0.15 = 0.015$ , and $1 - 0.015 = 0.985$ . (The complements of independent events are independent too, which is why $0.10 \cdot 0.15$ is allowed.)

A second detector raises the chance of an alarm from 0.90 to 0.985.

Worked exampleExample 3: exam-style free response

In a large town, 40% of households own a dog, 25% own a cat, and 55% own a dog or a cat (or both). Let $D$ = a randomly selected household owns a dog and $C$ = it owns a cat.

(a) Find the probability that a randomly selected household owns both a dog and a cat.

(b) Are $D$ and $C$ independent? Justify your answer.

(c) Two households are selected at random. Assuming the households are independent, find the probability that both own a dog or a cat.

Show solution

(a) Solve the addition rule for the joint probability:

P(D \cap C) = P(D) + P(C) - P(D \cup C) = 0.40 + 0.25 - 0.55 = 0.10.

The probability that a randomly selected household in the town owns both a dog and a cat is 0.10.

(b) $P(D) \cdot P(C) = 0.40 \cdot 0.25 = 0.10$ , which equals $P(D \cap C) = 0.10$ , so $D$ and $C$ are independent. Equivalently, $P(C \mid D) = \frac{0.10}{0.40} = 0.25 = P(C)$ : knowing a household has a dog doesn't change the probability that it has a cat.

P(\text{both households own a dog or a cat}) = 0.55 \cdot 0.55 = 0.3025.

Why these moves: (a) is the addition rule run backward, the one way to get a joint probability from a union. (b) needs a numerical comparison, so it shows both sides of $P(D \cap C) = P(D)P(C)$ and states the conclusion. (c) uses the stated independence of the households, which is different from the independence of $D$ and $C$ within one household.

Where students lose points

Check yourself

Question 1
Events $A$ and $B$ are independent, with $P(A) = 0.6$ and $P(B) = 0.3$ . What is $P(A \cup B)$ ?
1. A
  $0.18$
2. B
  $0.28$
3. C
  $0.72$
4. D
  $0.90$
Show answer
C
Independence gives $P(A \cap B) = 0.6 \cdot 0.3 = 0.18$ , and the addition rule gives $0.6 + 0.3 - 0.18 = 0.72$ . Choice D adds without removing the overlap, which counts the outcomes in both events twice.
Question 2
Events $A$ and $B$ are mutually exclusive, with $P(A) = 0.2$ and $P(B) = 0.5$ . Which statement is true?
1. A
  $A$ and $B$ are independent, because $P(A \cap B) = 0$ .
2. B
  $A$ and $B$ are not independent, because $P(A \cap B) = 0$ but $P(A) \cdot P(B) = 0.1$ .
3. C
  $P(A \cup B) = 0.6$
4. D
  $P(A \mid B) = 0.2$
Show answer
B
Independence needs $P(A \cap B) = P(A)P(B) = 0.1$ , but disjoint events have joint probability 0. Choice A confuses "can't happen together" with "don't affect each other"; in fact $P(A \mid B) = 0$ , not 0.2, and the union is $0.2 + 0.5 = 0.7$ .
Question 3
A student guesses at random on four multiple-choice questions, each with four choices, and the guesses are independent. What is the probability that the student gets at least one right?
Show answer
$1 - 0.75^4 \approx 0.684$
Each guess is wrong with probability $\frac{3}{4}$ . "None right" is four independent wrong guesses, $0.75^4 \approx 0.316$ , and "at least one" is its complement.
Question 4
In a group of 80 students, 20 are left-handed. The group's chess club has 25 members, and 5 of them are left-handed. For a student chosen at random from the 80, are "left-handed" and "in the chess club" independent? Justify.
Show answer
No. $P(\text{left-handed}) = \frac{20}{80} = 0.25$ , but $P(\text{left-handed} \mid \text{chess club}) = \frac{5}{25} = 0.20$ . The two differ, so the events are dependent.
Independence requires the conditional probability to equal the unconditional one exactly. Knowing a student is in the chess club lowers the probability of being left-handed from 0.25 to 0.20.

Independent Events and Unions of Events