Compound Probability Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
P(A)=0.3P(A)=0.3, P(B)=0.5P(B)=0.5, and A and B are mutually exclusive. Find P(A∪B)P(A \cup B).

Solution

  1. 1
    Mutually exclusive: P(A∩B)=0P(A \cap B) = 0
  2. 2
    P(A∪B)=P(A)+P(B)−P(A∩B)=0.3+0.5−0=0.8P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.5 - 0 = 0.8

Answer

P(A∪B)=0.8P(A \cup B) = 0.8
Mutually exclusive events cannot both occur simultaneously, so their intersection probability is 0. For mutually exclusive events, the addition rule simplifies to P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B). Note: mutually exclusive events are never independent (unless one has probability 0).

About Compound Probability

The probability of two or more events occurring together (P(A and B)P(A \text{ and } B)) or at least one occurring (P(A or B)P(A \text{ or } B)), accounting for whether the events are independent or dependent.

Learn more about Compound Probability →

More Compound Probability Examples

Example 1 medium

Events A and B: [formula], [formula], [formula]. Find (a) [formula], (b) [formula], and verify wheth

Example 2 hard

A card is drawn from a standard deck. Event A: card is red. Event B: card is a face card (J, Q, K).

Example 4 hard

Using the law of total probability: [formula], find [formula] given [formula], [formula], [formula].

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