AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/2/2-7)
Unit 2 · Topic 2.7
2.7 Independent Events and Unions of Events
Two events are independent when knowing one happened doesn't change the probability of the other. This topic gives you the tests for independence, the simple multiplication rule that follows from it and the addition rule for "A or B."
Key terms
- independent events
- union P(A ∪ B)
- addition rule
- multiplication rule for independent events
Independence
A and B are independent if P(A | B) = P(A). Learning that B happened gives no information about A. When that's true, P(B | A) = P(B) as well, and P(A ∩ B) = P(A) · P(B).
Any one of those three equations can be used to check independence. Compute both sides and compare.
- Coin flips, die rolls and draws with replacement are independent.
- Draws without replacement from a small group are not: the first draw changes what's left.
- Events in real data are rarely exactly independent. If the probabilities are clearly different, the events are dependent.
Checking with a two-way table
In the grade-and-sport table (200 students), P(sport) = 100/200 = 0.50 but P(sport | freshman) = 72/120 = 0.60. Those aren't equal, so for a randomly chosen student from this group, being a freshman and playing a sport are not independent.
Same check with the multiplication rule: P(freshman) · P(sport) = 0.60 × 0.50 = 0.30, but P(freshman ∩ sport) = 0.36. Not equal, so dependent.
The addition rule
P(A ∪ B), read "A union B," is the probability that A happens, B happens or both happen. The general addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B), and it's on the formula sheet.
You subtract the overlap because adding P(A) and P(B) counts the outcomes in both events twice. If the events are mutually exclusive, the overlap is 0 and the rule becomes P(A) + P(B).
Putting the rules together
Before you multiply, ask whether the events are independent. Before you add, ask whether they overlap. Those two questions prevent most probability mistakes.
Independence is often an assumption in a model, like assuming free throws or coin flips are independent. With real data from a two-way table, you check it by comparing conditional and overall proportions; small differences can come from chance, which is what the chi-square test in 3.14 is for.
- Not A: P(not A) = 1 − P(A).
- A and B, in general: P(A ∩ B) = P(A) · P(B | A).
- A and B, when independent: P(A ∩ B) = P(A) · P(B).
- A or B: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- A given B: P(A | B) = P(A ∩ B) ÷ P(B).
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Union from a table
From the grade-and-sport table, find the probability that a randomly chosen student is a freshman or plays a sport.
Show the solutionHide the solution
- Step 1: P(freshman) = 120/200 = 0.60. P(sport) = 100/200 = 0.50. P(freshman ∩ sport) = 72/200 = 0.36.
- Step 2: P(freshman ∪ sport) = 0.60 + 0.50 − 0.36 = 0.74.
- Step 3: Check by counting: everyone except seniors who don't play: 200 − 52 = 148, and 148/200 = 0.74.
Answer: 0.74
- Example 2Calculator allowed
Independent events and "at least one"
A 70% free-throw shooter takes two independent shots. Find P(makes both) and P(makes at least one).
Show the solutionHide the solution
- Step 1: Independent, so P(both) = 0.7 × 0.7 = 0.49.
- Step 2: At least one: use the complement. P(misses both) = 0.3 × 0.3 = 0.09.
- Step 3: P(at least one) = 1 − 0.09 = 0.91.
- Step 4: Check with the addition rule: 0.7 + 0.7 − 0.49 = 0.91.
Answer: P(both) = 0.49; P(at least one) = 0.91.
- Example 3Calculator allowed
Trap: assuming independence
In a class, 40% of students have a part-time job and 30% are on the honor roll. A student says P(job and honor roll) = 0.40 × 0.30 = 0.12. What is missing?
Show the solutionHide the solution
- Step 1: Multiplying probabilities is only valid if the events are independent.
- Step 2: Having a job could plausibly relate to honor roll status (less time for schoolwork), so independence can't be assumed.
- Step 3: You need either P(honor roll | job) or the joint probability from data.
Answer: The 0.12 assumes independence, which isn't given. Without P(honor roll | job) or a two-way table, you can't find the joint probability.
Common mistakes
- Multiplying P(A) · P(B) for "and" without checking that the events are independent.
- Forgetting to subtract P(A ∩ B) in the addition rule.
- Concluding independence because the events "seem unrelated" instead of checking probabilities.
- Calling mutually exclusive events independent.
On the exam
- "Are A and B independent? Justify." Compare P(A | B) with P(A), with numbers, and conclude.
- When a problem says trials are independent, it's telling you to multiply.
Connected topics
Videos
Check yourself
4 questions on 2.7 Independent Events and Unions of Events. Pick an answer to see if you got it, and why.
For two events, P(A) = 0.4, P(B) = 0.25 and P(A ∩ B) = 0.10. Which statement is true?
A commuter's train is on time 85% of the time, independently from day to day. What is the probability that the train is late at least once in 4 days?
In a town, 62% of households have a dog, 38% have a cat, and 20% have both. What percent of households have a dog or a cat?
Two machines in a factory work independently. Machine 1 breaks down on 5% of days and Machine 2 on 8% of days. On what percent of days do both machines break down?
0 of 4 answered