Conditional Probability, Independence and Bayes' Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Conditioning, independence and Bayes

Mathematics · Probability · ages 18-19
Name ______________________   Date ____________

  1. P(A and B) = 0.24 and P(B) = 0.6. Find P(A given B).

    Answer: ______________

  2. P(A and B) = 0.24 and P(B) = 0.6. Find P(A given B).

    Answer: ______________

  3. Which ratio is P(A given B)?

    • P(A and B) divided by P(B)
    • P(A and B) times P(B)
    • P(A or B) minus P(B)

  4. A and B are independent with P(B) above zero. Which of these equals P(A)?

    • P(A and B)
    • P(A given B)
    • P(A or B)

  5. A and B are mutually exclusive, each with positive chance. Mia says they must be independent. Is Mia right?

    Circle one:   True   False

  6. P(rain) = 0.3. P(late given rain) = 0.5 and P(late given no rain) = 0.2. Find P(late).

    • 0.7
    • 0.35
    • 0.29

  7. 10 percent of a high-risk group carry a condition. A test catches 90 percent of cases and clears 90 percent of healthy people. Given a positive result, what is the chance the person carries the condition?

    Answer: ______________

  8. 10% of a high-risk group has a condition. A test catches 90% of cases and correctly clears 90% of healthy people. Given a positive result, what is the chance the person has the condition?

    Answer: ______________

  9. A test is accurate and the condition is rare. Why can a positive result still probably be wrong?

    • The test misfires on every real case
    • Most positives come from healthy people, since the condition is rare
    • Healthy people never test positive

  10. A and B are independent with P(A) = 0.3 and P(B) = 0.4. Find P(A and B).

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Conditioning, independence and Bayes W1-mt_wIi_XRzwwK-s1

  1. 0.4 · Divide the joint chance by the marginal: 0.24 / 0.6 = 0.4.
  2. 0.4 · Divide the joint chance by the marginal: 0.24 divided by 0.6 is 0.4.
  3. P(A and B) divided by P(B) · Conditioning rescales the world to B, so divide the joint chance by P(B).
  4. P(A given B) · Independence means learning B changes nothing, so P(A given B) stays equal to P(A).
  5. False · Mia is wrong. Exclusivity forces P(A and B) = 0, while independence would need P(A) x P(B), which is positive. Both cannot hold.
  6. 0.29 · Split over weather: 0.3 times 0.5 plus 0.7 times 0.2 is 0.29.
  7. 0.5 · True positives are 0.09 and false positives are 0.09, so half of all positives are real: 0.09 divided by 0.18.
  8. 0.5 · True positives are 0.1 x 0.9 = 0.09 and false positives are 0.9 x 0.1 = 0.09, so half of all positives are real cases: 0.09 / 0.18 = 0.5.
  9. Most positives come from healthy people, since the condition is rare · Healthy people vastly outnumber cases, so their false positives can match or pass the true ones.
  10. 0.12 · Independence turns and into times: 0.3 times 0.4 is 0.12.
Worksheet · LightMySky