4.4 Mutually Exclusive Events

Two events are mutually exclusive if they cannot happen at the same time—formally P(A ∩ B) = 0. Quick checks you can use: - Think about the context: if A and B describe outcomes that can’t co-occur (e.g., “roll is 2” and “roll is 5”), they’re disjoint. - Use a Venn diagram: no overlap means mutually exclusive. - Use probabilities: if you calculate or are given P(A ∩ B) and it equals 0, they’re mutually exclusive. - Remember the addition rule for disjoint events: P(A ∪ B) = P(A) + P(B) only if they're mutually exclusive. Don’t confuse this with independence—independent events can both happen (P(A ∩ B) = P(A)P(B)), so independence ≠ mutual exclusivity (except in trivial cases). For more examples and AP-style practice on Topic 4.4, see the Fiveable study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and try practice problems (https://library.fiveable.me/practice/ap-statistics).

If A and B are mutually exclusive (disjoint), they can’t happen at the same time, so P(A ∩ B) = 0. That makes the addition rule simplify to P(A ∪ B) = P(A) + P(B) because the general rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (you’ll see this on the AP formula sheet). Quick tip: mutually exclusive ≠ independent. If events are mutually exclusive and one has positive probability, they can’t be independent (independence would require P(A ∩ B) = P(A)P(B) which would force one to be 0). For more examples and practice (aligned to the CED learning objective VAR-4.C), check the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and Unit 4 overview (https://library.fiveable.me/ap-statistics/unit-4). Want extra practice? Try the 1000+ AP Stats problems at (https://library.fiveable.me/practice/ap-statistics).

Use P(A ∩ B) = 0 when A and B are mutually exclusive (disjoint): they cannot happen at the same time, so their intersection is impossible. That’s exactly VAR-4.C.2 in the CED. Example: on one fair die roll, A = “even” and B = “odd” ⇒ P(A ∩ B) = 0. When it’s not zero, A and B can both occur together, so the intersection has some positive probability. Example: A = “multiple of 2” and B = “multiple of 3” on a six-sided die ⇒ A ∩ B = {6}, so P(A ∩ B) = 1/6. Quick tips for the exam: - If P(A ∩ B) = 0, you may use P(A ∪ B) = P(A) + P(B) (addition rule for disjoint events). - Don’t confuse mutually exclusive with independent: independent events can have P(A ∩ B) = P(A)P(B), often > 0, but they’re not disjoint unless one has probability 0. For more examples and practice, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and Unit 4 resources (https://library.fiveable.me/ap-statistics/unit-4). For lots of practice problems, go to (https://library.fiveable.me/practice/ap-statistics).

Mutually exclusive (disjoint) means the two events cannot happen at the same time—so P(A ∩ B) = 0. For disjoint A and B the addition rule simplifies to P(A ∪ B) = P(A) + P(B). Independent means the occurrence of one event gives no information about the other: P(A | B) = P(A). Equivalently, P(A ∩ B) = P(A)·P(B). Key difference: disjointness is about an impossible intersection; independence is about no change in probability when you know the other event happened. They’re not the same: if A and B are both disjoint AND both have positive probability, they cannot be independent (because P(A ∩ B)=0 ≠ P(A)P(B) if P(A),P(B)>0). The only way disjoint events can be independent is if one event has probability 0. This is exactly what Topic 4.4 (VAR-4.C) tests—review the CED definitions and examples in the Fiveable study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and practice problems at (https://library.fiveable.me/practice/ap-statistics).

Yes—joint probability and the intersection are the same idea. The joint probability P(A and B) is written P(A ∩ B) and means “the probability that A and B both occur” (CED VAR-4.C.1). If A and B are mutually exclusive (disjoint), they cannot happen together, so P(A ∩ B) = 0 (CED VAR-4.C.2). Quick reminders that often trip people up: joint/intersection ≠ conditional probability (P(A|B) = P(A ∩ B)/P(B)), and mutually exclusive is different from independent—independent events can both occur (so P(A ∩ B) = P(A)P(B)), while mutually exclusive events cannot. For more on this AP topic, check the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and practice problems (https://library.fiveable.me/practice/ap-statistics).

Step-by-step: how to tell if two events are disjoint (mutually exclusive) 1. Translate the events into outcomes in the sample space S. Write A and B as sets of outcomes (or list them). 2. Find the intersection A ∩ B—the outcomes that would make both A and B happen at the same time. Use a Venn diagram if that helps. 3. Check P(A ∩ B). If P(A ∩ B) = 0 (no outcomes in common), then A and B are disjoint/mutually exclusive. If P(A ∩ B) > 0, they are not disjoint. (This is the CED rule: VAR-4.C.2.) 4. Optional quick checks: if the wording makes simultaneous occurrence impossible (e.g., “rolling an even number” and “rolling an odd number” on one die), they’re disjoint. But don’t confuse disjoint with independent—disjoint events cannot be independent unless one has probability 0. 5. Use the addition rule for disjoint events: if disjoint, P(A ∪ B) = P(A) + P(B). Otherwise use P(A ∪ B) = P(A)+P(B) − P(A ∩ B). For more examples and practice aligned to the AP CED, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and lots of practice questions (https://library.fiveable.me/practice/ap-statistics).

Take each event as a set of outcomes from the die (sample space S = {1,2,3,4,5,6}). Two events are mutually exclusive iff they cannot happen at the same time, i.e. P(A ∩ B) = 0 (CED VAR-4.C.2). Steps: 1. Write the events as outcome sets. Example: A = “even” = {2,4,6}; B = “4” = {4}. 2. Find the intersection A ∩ B. If it’s empty, they’re mutually exclusive; if not, they aren’t. - A ∩ B = {4} ≠ ∅ → not mutually exclusive. - Example: C = “odd” = {1,3,5}. C ∩ A = ∅ → C and A are mutually exclusive. 3. Use P(A ∩ B) = 0 to justify your answer on the exam; if mutually exclusive, P(A ∪ B) = P(A)+P(B) (addition rule). Remember: mutually exclusive ≠ independent. Two mutually exclusive non-impossible events are dependent (their probabilities affect each other). For more examples and quick practice, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and Unit 4 review (https://library.fiveable.me/ap-statistics/unit-4). For extra practice problems, try the AP practice bank (https://library.fiveable.me/practice/ap-statistics).

P(A ∩ B) is the joint probability that both events A and B occur—basically the probability of their intersection (what’s in both sets). In AP language it’s the probability that A and B happen at the same time (VAR-4.C.1). P(A ∩ B) = 0 exactly when A and B are mutually exclusive (disjoint): they cannot happen together, so their intersection is the empty set (VAR-4.C.2). Example: on a single die roll, A = “even” and B = “odd” → P(A ∩ B) = 0 because no outcome is both even and odd. Note: mutually exclusive is different from independent—independent events can have P(A ∩ B) = P(A)P(B) (not necessarily zero). For more on this topic see the Fiveable study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and the Unit 4 overview (https://library.fiveable.me/ap-statistics/unit-4). For extra practice, try problems at (https://library.fiveable.me/practice/ap-statistics).

Mutually exclusive means two events can’t happen at the same time, so P(A ∩ B) = 0. Simple examples you can picture: - Tossing one fair coin: A = “heads”, B = “tails.” You can’t get both on one toss. - Rolling one die: A = “roll is even” (2,4,6), B = “roll is odd” (1,3,5). A ∩ B is impossible. - Drawing one card: A = “card is an Ace,” B = “card is a King.” One card can’t be both. - Class outcome: A = “student passes,” B = “student fails” (assuming no partial outcomes). Use a Venn diagram to see disjoint circles with no overlap. Remember: mutually exclusive ≠ independent—independent events can occur together. On the exam you’ll use P(A ∪ B) = P(A) + P(B) when events are disjoint (Topic 4.4, VAR-4.C). For a quick review, check the Fiveable study guide for mutually exclusive events (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6). For more unit review and lots of practice problems, see the Unit 4 page (https://library.fiveable.me/ap-statistics/unit-4) and practice bank (https://library.fiveable.me/practice/ap-statistics).

Write P(A and B) = 0 whenever A and B cannot happen at the same time—i.e., their intersection is impossible. That's exactly what "mutually exclusive" (disjoint) means in the CED: P(A ∩ B) = 0 (VAR-4.C.2). Quick checks you can use on a problem: - Ask: can one single outcome satisfy both A and B? If no, set P(A and B)=0. - Use a Venn diagram: if the circles don't overlap, intersection area = 0. - Use logic/context: A = “roll a 2” and B = “roll a 3” → P(A and B)=0; A = “even” and B = “odd” → 0. Don’t confuse this with independence: two events can be independent and still have P(A ∩ B) ≠ 0. If the problem gives probabilities that lead to a nonzero intersection, then the events are not mutually exclusive. For extra practice and examples, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and unit resources (https://library.fiveable.me/ap-statistics/unit-4).

They're the same thing: “disjoint” and “mutually exclusive” both mean two events cannot happen at the same time. In CED language (Topic 4.4, VAR-4.C), that means P(A ∩ B) = 0. So if A and B are disjoint/mutually exclusive, the addition rule simplifies to P(A ∪ B) = P(A) + P(B). Two quick reminders that show common confusion: - Mutually exclusive ≠ independent. If A and B are mutually exclusive and one has positive probability, knowing A happened changes the chance B happens (to 0), so they’re usually not independent. - “Overlapping” events are NOT disjoint; they have P(A ∩ B) > 0. If you want a short CED-aligned read and examples, check the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6). For extra practice, Fiveable has lots of practice problems (https://library.fiveable.me/practice/ap-statistics).

Say the definition first, then show it fails. Start with the CED idea: two events are mutually exclusive iff P(A ∩ B) = 0 (they can’t occur together). In a free-response answer you should: - State the definition: “Mutually exclusive means P(A ∩ B) = 0 (no shared outcomes).” - Produce evidence that A and B can occur together: give a specific outcome in the sample space that lies in both A and B (or draw/describe a Venn diagram with an overlapping region). - Use probability values if given: compute or point out P(A ∩ B) > 0 (or show P(A ∪ B) ≠ P(A)+P(B) unless you subtract an intersection). - Conclude in context: “Because outcome x is in both A and B (so P(A ∩ B) > 0), A and B are not mutually exclusive.” That’s all AP wants: definition + a concrete counterexample or a computed positive intersection and a context sentence. For extra practice, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and more problems at (https://library.fiveable.me/practice/ap-statistics).

Ask: can A and B happen at the same time? If yes, they’re not mutually exclusive; if no, they are. In card terms: “A = draw a king” and “B = draw a heart.” These are NOT mutually exclusive because a card can be the king of hearts (P(A ∩ B) > 0). But “A = draw a king” and “C = draw a queen” ARE mutually exclusive—one card can’t be both a king and a queen, so P(A ∩ C) = 0. Use the intersection test from the CED: compute or reason about P(A ∩ B). If P(A ∩ B) = 0 the events are disjoint/mutually exclusive and you can add probabilities directly: P(A ∪ B) = P(A)+P(B). If P(A ∩ B) > 0 they overlap and you must use P(A ∪ B)=P(A)+P(B)−P(A ∩ B). For quick review, check the Topic 4.4 study guide (Fiveable) (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and practice problems at (https://library.fiveable.me/practice/ap-statistics).

Your textbook writes P(A ∩ B) = 0 only when A and B are mutually exclusive (disjoint)—meaning they cannot happen at the same time (CED VAR-4.C.2). In that case the intersection is impossible, so P(A ∩ B) = 0. Example: on one die roll A = “even” and B = “odd” → intersection empty → P(A ∩ B) = 0. If A and B can occur together, P(A ∩ B) > 0. Example: A = “multiple of 2” and B = “multiple of 3” on a die → both can happen (6), so the intersection isn’t empty. Use Venn diagrams or the sample space to check whether any outcome is in both events. Quick reminders for AP exam: use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (CED formula) and don’t confuse mutually exclusive with independent—mutually exclusive events with positive probabilities cannot be independent. For a short refresher, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6). For more practice, try problems at (https://library.fiveable.me/practice/ap-statistics).

Short answer: only in a trivial case. Mutually exclusive (disjoint) means P(A ∩ B) = 0 (CED VAR-4.C.2). Independence means P(A ∩ B) = P(A)P(B). So if both hold, P(A)P(B) = 0, which means at least one of P(A) or P(B) must equal 0. In other words, two events with positive probability cannot be both mutually exclusive and independent. Example: A = impossible event (P(A)=0) and any B—then A and B are disjoint and independent, but that’s a degenerate case. This idea shows up on AP questions about independence vs. disjointness (Topic 4.4 and 4.6). For a quick review, see the Topic 4.4 study guide (https://library.fiveable.me/ap-statistics/unit-4/mutually-exclusive-events/study-guide/iBljImMDLJ8bSWeWXbi6) and more Unit 4 resources (https://library.fiveable.me/ap-statistics/unit-4). For extra practice, check the practice problem bank (https://library.fiveable.me/practice/ap-statistics).