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

Conditioning, independence and Bayes

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

  1. In a club, P(a member plays chess) = 0.4. Of the chess players, 0.5 also play piano. Find P(chess and piano).

    Answer: ______________

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

    Answer: ______________

  3. P(B) > 0. Which ratio is P(A given B)?

    • P(A and B) divided by P(B)
    • P(A and B) multiplied by P(B) instead
    • P(A or B) with P(B) taken away from it

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

    • P(A and B), because both events must happen together
    • P(A given B), because learning B changes nothing
    • P(A or B), because it takes in every outcome of A

  5. P(A) = 0.5, P(B) = 0.2 and P(A and B) = 0.1. Kim says A and B are independent. Is Kim 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. 20 percent of a group carry a condition. A test catches 60 percent of carriers and wrongly flags 5 percent of everyone else. Given a positive result, what is the chance the person carries the condition?

    Answer: ______________

  8. Out of 1000 people, 100 carry a condition. A test catches 80 of the 100 carriers and wrongly flags 20 of the 900 others. Given a positive result, what is the chance the person carries the condition? Give a decimal.

    Answer: ______________

  9. A condition is rare and a test catches most carriers. Why might a positive result still be more likely false than true?

    • Every positive result must be false whenever a tested condition is rare
    • False positives in the larger healthy group can outnumber true ones
    • The chance of a true positive depends only on sensitivity and never on the base rate

  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. 0.2 · Along the path, multiply the first chance by the conditional chance: 0.4 times 0.5 = 0.2.
  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) · Since P(B) > 0, divide the joint chance P(A and B) by P(B) to find the chance of A within B.
  4. P(A given B), because learning B changes nothing · Independence means learning B changes nothing, so P(A given B) stays equal to P(A).
  5. True · Kim is right. For independent events P(A and B) = P(A) times P(B), and 0.5 times 0.2 = 0.1.
  6. 0.29 · Split over weather: 0.3 times 0.5 plus 0.7 times 0.2 is 0.29.
  7. 0.75 · True positives are 0.2 times 0.6 = 0.12 and false positives are 0.8 times 0.05 = 0.04, so 0.12 divided by 0.16 is 0.75.
  8. 0.8 · True positives are 80 and false positives are 20, so 80 of the 100 positives are real: 80 divided by 100 = 0.8.
  9. False positives in the larger healthy group can outnumber true ones · The healthy group is much larger, so its false positives can outnumber the true positives among carriers. Whether this happens depends on the base rate, sensitivity and false-positive rate.
  10. 0.12 · Independence turns and into times: 0.3 times 0.4 is 0.12.
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