Compound Probability Math Example 1

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

Example 1

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Events A and B: P(A)=0.5P(A)=0.5, P(B)=0.4P(B)=0.4, P(A∩B)=0.2P(A \cap B)=0.2. Find (a) P(A∪B)P(A \cup B), (b) P(A∣B)P(A|B), and verify whether A and B are independent.

Solution

  1. 1
    (a) Addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)=0.5+0.4−0.2=0.7P(A \cup B) = P(A)+P(B)-P(A \cap B) = 0.5+0.4-0.2 = 0.7
  2. 2
    (b) Conditional probability: P(A∣B)=P(A∩B)P(B)=0.20.4=0.5P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5
  3. 3
    Independence check: P(A∣B)=0.5=P(A)P(A|B) = 0.5 = P(A) → A and B are independent
  4. 4
    Verify: P(A)×P(B)=0.5×0.4=0.20=P(A∩B)P(A) \times P(B) = 0.5 \times 0.4 = 0.20 = P(A \cap B) ✓

Answer

(a) P(A∪B)=0.7P(A \cup B) = 0.7. (b) P(A∣B)=0.5P(A|B) = 0.5. A and B are independent.
Compound probability uses the addition rule for unions and the multiplication rule for intersections. When P(A∣B)=P(A)P(A|B) = P(A), the events are independent. The addition rule prevents double-counting the intersection: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).

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 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 3 easy

[formula], [formula], and A and B are mutually exclusive. Find [formula].

Example 4 hard

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

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