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Bayes Theorem worksheet

Grades 11–12 · CCSS HSS.CP.B.8, HSS.CP.B.6, HSS.CP.A.3

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 30%
- P(B | A), probability of B when A is true: 85%
- P(B), overall probability of B: 48%
Find: P(A | B)
Answer
2.
Problem 2.
- Enter: Base rate and test accuracy
- Base rate: how common the condition is, P(A): 1%
- True positive rate (sensitivity), P(+ | A): 95%
- False positive rate, P(+ | not A): 4%
Find: Chance of the condition after a positive result, P(A | +)
Answer
3.
Problem 3.
- Enter: Base rate and test accuracy
- Base rate: how common the condition is, P(A): 1%
- True positive rate (sensitivity), P(+ | A): 90%
- False positive rate, P(+ | not A): 6%
Find: Chance of the condition after a positive result, P(A | +)
Answer
4.
Problem 4.
- Enter: Base rate and test accuracy
- Base rate: how common the condition is, P(A): 1%
- True positive rate (sensitivity), P(+ | A): 78%
- False positive rate, P(+ | not A): 5%
Find: Chance of the condition after a positive result, P(A | +)
Answer
5.
Problem 5.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 28%
- P(B | A), probability of B when A is true: 70%
- P(B), overall probability of B: 45%
Find: P(A | B)
Answer
6.
Problem 6.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 34%
- P(B | A), probability of B when A is true: 80%
- P(B), overall probability of B: 38%
Find: P(A | B)
Answer
7.
Problem 7.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 32%
- P(B | A), probability of B when A is true: 76%
- P(B), overall probability of B: 38%
Find: P(A | B)
Answer
8.
Problem 8.
- Enter: Base rate and test accuracy
- Base rate: how common the condition is, P(A): 1%
- True positive rate (sensitivity), P(+ | A): 90%
- False positive rate, P(+ | not A): 5%
Find: Chance of the condition after a positive result, P(A | +)
Answer
9.
Problem 9.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 26%
- P(B | A), probability of B when A is true: 100%
- P(B), overall probability of B: 36%
Find: P(A | B)
Answer
10.
Problem 10.
- Enter: P(A), P(B | A) and P(B)
- P(A), probability of A: 20%
- P(B | A), probability of B when A is true: 85%
- P(B), overall probability of B: 55%
Find: P(A | B)
Answer

Answer key: Bayes Theorem worksheet

1. P(A | B): 53.13%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 0.85 \times 0.3$ $= 0.255$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.255}{0.48}$ $= 0.53125$ $= 53.125%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.48 - 0.255}{1 - 0.3}$ $= 0.3214285714$
2. Chance of the condition after a positive result, P(A | +): 19.35%
1. Find the chance of having the condition and testing positive: how often the test catches it, times how common it is
  $P(A and +)$ $= 0.95 \times 0.01$ $= 0.0095$
2. Find the chance of a false alarm: the false positive rate, times the share of people without the condition
  $P(not A and +)$ $= 0.04 \times 1 - 0.01$ $= 0.0396$
3. Add the two, since a positive result can come either way. That is the total chance of testing positive
  $P(+)$ $= 0.0095 + 0.0396$ $= 0.0491$
4. Divide the real positives by all positives to see how many positive results are real (Bayes' theorem)
  $P(A | +)$ $= \frac{0.0095}{0.0491}$ $= 0.1934826884$ $= 19.348269%$
5. After a negative result: divide the missed cases by all negative results
  $P(A | −)$ $= \frac{0.0005}{0.9509}$ $= 0.0005258176$
3. Chance of the condition after a positive result, P(A | +): 13.16%
1. Find the chance of having the condition and testing positive: how often the test catches it, times how common it is
  $P(A and +)$ $= 0.9 \times 0.01$ $= 0.009$
2. Find the chance of a false alarm: the false positive rate, times the share of people without the condition
  $P(not A and +)$ $= 0.06 \times 1 - 0.01$ $= 0.0594$
3. Add the two, since a positive result can come either way. That is the total chance of testing positive
  $P(+)$ $= 0.009 + 0.0594$ $= 0.0684$
4. Divide the real positives by all positives to see how many positive results are real (Bayes' theorem)
  $P(A | +)$ $= \frac{0.009}{0.0684}$ $= 0.1315789474$ $= 13.157895%$
5. After a negative result: divide the missed cases by all negative results
  $P(A | −)$ $= \frac{0.001}{0.9316}$ $= 0.0010734221$
4. Chance of the condition after a positive result, P(A | +): 13.61%
1. Find the chance of having the condition and testing positive: how often the test catches it, times how common it is
  $P(A and +)$ $= 0.78 \times 0.01$ $= 0.0078$
2. Find the chance of a false alarm: the false positive rate, times the share of people without the condition
  $P(not A and +)$ $= 0.05 \times 1 - 0.01$ $= 0.0495$
3. Add the two, since a positive result can come either way. That is the total chance of testing positive
  $P(+)$ $= 0.0078 + 0.0495$ $= 0.0573$
4. Divide the real positives by all positives to see how many positive results are real (Bayes' theorem)
  $P(A | +)$ $= \frac{0.0078}{0.0573}$ $= 0.1361256545$ $= 13.612565%$
5. After a negative result: divide the missed cases by all negative results
  $P(A | −)$ $= \frac{0.0022}{0.9427}$ $= 0.0023337223$
5. P(A | B): 43.56%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 0.7 \times 0.28$ $= 0.196$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.196}{0.45}$ $= 0.4355555556$ $= 43.555556%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.45 - 0.196}{1 - 0.28}$ $= 0.3527777778$
6. P(A | B): 71.58%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 0.8 \times 0.34$ $= 0.272$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.272}{0.38}$ $= 0.7157894737$ $= 71.578947%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.38 - 0.272}{1 - 0.34}$ $= 0.1636363636$
7. P(A | B): 64%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 0.76 \times 0.32$ $= 0.2432$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.2432}{0.38}$ $= 0.64$ $= 64%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.38 - 0.2432}{1 - 0.32}$ $= 0.2011764706$
8. Chance of the condition after a positive result, P(A | +): 15.38%
1. Find the chance of having the condition and testing positive: how often the test catches it, times how common it is
  $P(A and +)$ $= 0.9 \times 0.01$ $= 0.009$
2. Find the chance of a false alarm: the false positive rate, times the share of people without the condition
  $P(not A and +)$ $= 0.05 \times 1 - 0.01$ $= 0.0495$
3. Add the two, since a positive result can come either way. That is the total chance of testing positive
  $P(+)$ $= 0.009 + 0.0495$ $= 0.0585$
4. Divide the real positives by all positives to see how many positive results are real (Bayes' theorem)
  $P(A | +)$ $= \frac{0.009}{0.0585}$ $= 0.1538461538$ $= 15.384615%$
5. After a negative result: divide the missed cases by all negative results
  $P(A | −)$ $= \frac{0.001}{0.9415}$ $= 0.0010621349$
9. P(A | B): 72.22%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 1 \times 0.26$ $= 0.26$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.26}{0.36}$ $= 0.7222222222$ $= 72.222222%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.36 - 0.26}{1 - 0.26}$ $= 0.1351351351$
10. P(A | B): 30.91%
1. Turn the percents into decimals. Then multiply to find the chance that A and B both happen
  $P(A and B)$ $= 0.85 \times 0.2$ $= 0.17$
2. Divide by the chance of B to find how often A is true among the cases where B happens (Bayes' theorem)
  $P(A | B)$ $= \frac{0.17}{0.55}$ $= 0.3090909091$ $= 30.909091%$
3. Work out how often B happens without A: the rest of B's chance, spread over the cases without A
  $P(B | not A)$ $= \frac{0.55 - 0.17}{1 - 0.2}$ $= 0.475$