Compound Probability Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Compound Probability.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

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

Single-event probability asks about one thing happening. Compound probability asks about combinations: 'What's the chance of rolling a 6 AND flipping heads?' or 'What's the chance of drawing a heart OR a face card?' The word 'and' usually means multiply; the word 'or' usually means add (but subtract the overlap).

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: For 'and,' multiply probabilities (adjusting for dependence). For 'or,' add probabilities and subtract the overlap to avoid double-counting.

Common stuck point: The 'or' formula requires subtracting P(A \text{ and } B) to avoid counting the overlap twice. If events are mutually exclusive (can't both happen), the overlap is zero.

Worked Examples

Example 1

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

Solution

  1. 1
    (a) Addition rule: P(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) = \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) → A and B are independent
  4. 4
    Verify: P(A) \times P(B) = 0.5 \times 0.4 = 0.20 = P(A \cap B) ✓

Answer

(a) P(A \cup B) = 0.7. (b) P(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), the events are independent. The addition rule prevents double-counting the intersection: P(A \cup B) = P(A) + P(B) - P(A \cap B).

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). Find P(A \cup B) and P(A \cap B).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

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

Example 2

hard
Using the law of total probability: P(B) = P(B|A)P(A) + P(B|A^c)P(A^c), find P(B) given P(A)=0.4, P(B|A)=0.7, P(B|A^c)=0.3.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

probabilityindependent eventsconditional probability