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Compound Probability

A free Pre-Algebra lesson from the “Statistics and Probability” unit, with a worked example and practice problems including step-by-step solutions.

Two events are independent if one does not affect the other (like rolling a die and flipping a coin). For independent events, multiply: P(A and B) = P(A) * P(B). For mutually exclusive events (can't both happen), add: P(A or B) = P(A) + P(B). For 'at least one' problems, use the complement. In Statistics and Probability, the goal is not just to get an answer but to recognize the structure of the problem, choose a reliable strategy, and explain why the result is reasonable. The practice set now includes targeted skill work, transfer questions, and mixed review so students build fluency and retention.

What you'll learn

Why it matters: Card games, dice games, weather forecasts, and reliability engineering (the chance two safety systems both fail) all use compound probabilities.

Worked example

Problem. Flip a coin and roll a die. Find P(heads AND 6) as a fraction.

  1. P(heads) = 1/2 and P(6) = 1/6.
  2. Independent events multiply: (1/2)(1/6) = 1/12.
  3. Connect the calculation back to Compound Probability so the method, not just the arithmetic, is clear.
  4. Check the result against the original question before writing the final answer.

Answer: 1/12

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (typed response). Counts include repeated prompts and variants.

Practice problems

1. Flip a coin twice. P(heads, heads) as a fraction.

Show solution
  1. Each flip: P(H) = 1/2.
  2. Multiply: (1/2)(1/2) = 1/4.

Answer: 1/4

2. Roll two dice. P(both 6) as a fraction.

Show solution
  1. Each die: P(6) = 1/6.
  2. (1/6)(1/6) = 1/36.

Answer: 1/36

3. Flip a coin and roll a die. P(tails AND 4) as a fraction.

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  1. (1/2)(1/6) = 1/12.

Answer: 1/12

4. Roll two dice. P(sum equals 7) as a fraction.

Show solution
  1. 6 ways out of 36: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
  2. 6/36 = 1/6.

Answer: 1/6

5. Flip a coin twice. P(at least one heads) as a fraction.

Show solution
  1. P(no heads) = (1/2)(1/2) = 1/4.
  2. P(at least one) = 1 - 1/4 = 3/4.

Answer: 3/4

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