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Mutually Exclusive Events

Mutually Exclusive Events – definition and example, Algebra 2 math glossary – Mathovore

Key takeaway: Mutually exclusive events are two events that cannot both happen in the same trial, so they share no outcomes and P(A and B) = 0.

Grade 11 – Algebra 2S-CP.B.7Probability and Counting

Definition of Mutually Exclusive Events

Definition

Two events are mutually exclusive events when the occurrence of one rules out the other. A single draw cannot satisfy both. Rolling a die and getting a 2 excludes getting a 6 on that same roll, and one student cannot be born in both July and November.

Picture them as two circles in a Venn diagram that never touch. No outcome lies in both, so the overlap is empty and its probability is zero. Such events are also called disjoint. Knowing that two events are exclusive makes adding their probabilities safe, because nothing is counted twice.

Formula and notation

Events \(A\) and \(B\) are mutually exclusive exactly when

\[P(A\text{ and }B)=0 \quad\text{so}\quad P(A\text{ or }B)=P(A)+P(B).\]

In set language, \(A\cap B=\varnothing\), the empty set. An event and its complement are always exclusive.

A: multiple of 77, 14B: over 1718, 19, 20no overlaptickets numbered 1 to 20
Circles that do not touch show events with no common outcome.

Key facts

  • Test for mutually exclusive events: list each outcome; if none appears in both lists, they qualify.
  • Exclusive events with positive probabilities are never independent of each other.
  • The probabilities of all exclusive events in a full list add up to 1.
  • Being exclusive depends on the trial: two events may overlap in a different experiment.

Mutually Exclusive Events example

Problem

A raffle drum holds tickets numbered 1 to 20. Event A is drawing a multiple of 7. Event B is drawing a number greater than 17. Show that A and B are exclusive and find \(P(A\text{ or }B)\).

  1. List A: 7 and 14. List B: 18, 19 and 20. The lists share nothing.

  2. Probabilities: \(P(A)=\frac{2}{20}\) and \(P(B)=\frac{3}{20}\).

  3. Add them: \(\frac{2}{20}+\frac{3}{20}=\frac{5}{20}=\frac{1}{4}\).

The events are exclusive, and \(P(A\text{ or }B)=\)\(\frac{1}{4}\).

Common mistake

Watch out

Zora says that because the events above cannot happen together, they must be independent.

Why it is wrong

Mutually exclusive events are strongly dependent: if A happens, B is certain not to. Independence would need \(P(A\text{ and }B)=P(A)P(B)=\frac{2}{20}\cdot\frac{3}{20}\neq 0\), but here the overlap is 0.

Related terms and lessons

  • Algebra 2 · Probability: Sample Spaces, Independence and Conditional ProbabilityLessonWorksheetQuiz

Browse the glossary: A–Z list · Algebra 2 vocabulary · Probability and Counting

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