Independent Events and Unions of Events

Part 1 · Hook

Why this matters

A phone reviewer tests two brands and finds that 60% of each brand's batteries last a full day. Does knowing the brand help you predict the battery? Not at all, and that is what independence means. A Venn diagram of students in two classes shows the other half of this topic: how to add probabilities that overlap.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Of 40 art students, 15 take Spanish. What is P(Spanish given art)?

0.375
0.15
0.3
0.125

Show the answer

Given art, divide by the 40 art students: $\frac{15}{40} = 0.375$ .

Correct: 0.375:
0.15:
0.3:
0.125:

2. Events R and T are mutually exclusive. What is $P(R \cap T)$ ?

0
1
$P(R) + P(T)$
0.5

Show the answer

Mutually exclusive events cannot happen together.

Correct: 0:
1:
$P(R) + P(T)$ :
0.5:

Part 4 · See it

See it first

Venn diagram of 120 students: 50 take Spanish, 40 take art, and 15 take both, so 35 take only Spanish, 25 only art and 45 neither. The union, Spanish or art, is shaded: 35 + 15 + 25 = 75 students, which equals 50 + 40 − 15. — 120 students: 50 take Spanish, 40 take art, 15 take both. The shaded union holds 75 students: 50 + 40 − 15. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Knowing that B happened leaves the probability of A unchangedA and B are independent: $P(A \mid B) = P(A)$
Put $P(A \mid B) = P(A)$ into the multiplication rulefor independent events, $P(A \cap B) = P(A) \cdot P(B)$
Add P(A) and P(B) for "A or B"the outcomes in both are counted twice
Subtract the overlap oncethe general addition rule: $P(A \cup B)$ $= P(A) + P(B) - P(A \cap B)$

Part 6 · Key ideas

Key ideas

Events are independent when knowing one happened does not change the probability of the other: $P(A \mid B) = P(A)$ . Then, and only then, $P(A \cap B) = P(A) \cdot P(B)$ .
To check independence, compare $P(A \mid B)$ with $P(A)$ , or $P(A \cap B)$ with $P(A) \cdot P(B)$ , using the numbers.
The union $A \cup B$ is "A or B or both". The general addition rule: $P(A \cup B)$ $= P(A) + P(B) - P(A \cap B)$ .
A Venn diagram shows events as overlapping circles; fill the overlap first.
Mutually exclusive is not independent: events that cannot happen together affect each other completely.

Part 7 · Misconception

A common mistake

The wrong idea: Mutually exclusive events are independent, because one has nothing to do with the other.

What actually happens: If two events cannot happen together, knowing one happened makes the other impossible, so its probability drops to 0. They are as dependent as events can be.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Spanish and art students

At a school, 120 students are in the arts and languages program. 50 take Spanish, 40 take art, and 15 take both. One of the 120 students is chosen at random (invented data).

1. What is the probability that the student takes Spanish or art?

0.75
0.125
0.625
0.375

Show the answer

Addition rule: $P(S \cup R)$ $= P(S)$ $+ P(R) - P(S \cap R)$ $= \frac{50}{120}$ $+ \frac{40}{120} - \frac{15}{120}$ $= \frac{75}{120}$ $= 0.625$ , where S is Spanish and R is art.

0.75: That is $\frac{90}{120}$ : the 15 students in both were counted twice.
0.125: That is $\frac{15}{120}$ , the students who take both, not either.
Correct: 0.625: Right: 75 of the 120 students take at least one of the two.
0.375: That is $\frac{45}{120}$ , the students who take neither.

2. Given that the student takes art, what is the probability that the student takes Spanish?

0.3
0.125
0.417
0.375

Show the answer

Given art, the 40 art students are the total, and 15 of them take Spanish: $\frac{15}{40} = 0.375$ .

0.3: That is $\frac{15}{50}$ , the probability of art given Spanish.
0.125: That is $\frac{15}{120}$ , the joint probability.
0.417: That is $\frac{50}{120}$ , the probability of Spanish for all students.
Correct: 0.375: Right: 15 of the 40 art students.

3. Are "takes Spanish" and "takes art" independent events?

No: P(Spanish given art) is 0.375 but P(Spanish) is about 0.417
Yes, because 15 students take both classes, so the events overlap
No, because they are mutually exclusive events at this school
Yes, because the students chose their own classes

Show the answer

Compare $P(\text{Spanish} \mid \text{art})$ $= \frac{15}{40}$ $= 0.375$ with $P(\text{Spanish})$ $= \frac{50}{120}$ $\approx 0.417$ . They differ, so the events are not independent.

Correct: No: P(Spanish given art) is 0.375 but P(Spanish) is about 0.417: Right: knowing art changes the probability of Spanish.
Yes, because 15 students take both classes, so the events overlap: Overlapping is not independence; independence compares probabilities.
No, because they are mutually exclusive events at this school: They are not mutually exclusive: 15 students take both.
Yes, because the students chose their own classes: How classes were chosen does not settle it; compare the probabilities.

Table

Phone brand and battery life

A tech site tested a random sample of 250 phones of two brands and recorded whether the battery lasted a full day of normal use. One of the 250 phones is chosen at random (invented data).

Brand and battery result for 250 phones
Brand	Lasted a full day	Did not	Total
Brand X	60	40	100
Brand Y	90	60	150
Total	150	100	250

4. Are "Brand Y" and "battery lasted a full day" independent?

No: more Brand Y phones (90) than Brand X phones (60) lasted a full day
No: a phone can be Brand Y and last a full day, so they overlap
Yes: a phone is either Brand X or Brand Y and is not both at once
Yes: the probability of a full day is 0.6 with or without knowing the brand

Show the answer

$P(\text{full day} \mid \text{Brand Y})$ $= \frac{90}{150}$ $= 0.6$ , the same as $P(\text{full day}) = 0.6$ (and Brand X is $\frac{60}{100} = 0.6$ too). Knowing the brand does not change the probability, so the events are independent.

No: more Brand Y phones (90) than Brand X phones (60) lasted a full day: Counts depend on how many of each brand were tested; compare probabilities.
No: a phone can be Brand Y and last a full day, so they overlap: Overlap is not how independence is checked.
Yes: a phone is either Brand X or Brand Y and is not both at once: That makes the brands mutually exclusive, which is a different idea.
Correct: Yes: the probability of a full day is 0.6 with or without knowing the brand: Right: equal conditional and unconditional probabilities.

5. A factory has two machines that work independently. On a given day, machine 1 breaks down with probability 0.05 and machine 2 with probability 0.08. What is the probability that both break down on the same day?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

For independent events, $P(\text{both})$ $= (0.05)(0.08)$ $= 0.004$ .

Answer: 0.004

6. In a town, 55% of households have a streaming subscription, 30% have cable, and 15% have both. What is the probability that a randomly chosen household has streaming or cable?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

General addition rule: $0.55 + 0.30 - 0.15$ $= 0.70$ . Subtract the 15% with both, who were counted in each group.

Answer: 0.7

Part 9 · Summary

Summary

Events are independent when

P(A \mid B) = P(A)

; then the probability of both is the product

P(A) \cdot P(B)

. Check independence with the numbers, never by assumption alone. The probability of A or B is

P(A) + P(B) - P(A \cap B)

, subtracting the overlap that would be counted twice; for mutually exclusive events the overlap is 0.

Part 10 · Up next

What comes next

Part 11 · Connections