A diagnostic model relates a dental cavity to the symptom toothache. The…
2021
A diagnostic model relates a dental cavity to the symptom toothache. The following probabilities are given: P(cavity) = 0.2, P(toothache | cavity) = 0.6, and P(toothache | ¬cavity) = 0.1.
Using these values, the probability of a cavity given evidence of a toothache, P(cavity | toothache), is __________.
- A.
0.400
- B.
0.600
- C.
0.280
- D.
0.216
Attempted by 14 students.
Show answer & explanation
Correct answer: B
Concept
Bayes’ rule recovers a posterior probability from a prior and a likelihood: P(A | B) = P(B | A)·P(A) / P(B). The denominator P(B) is the total probability of the evidence and is expanded over all mutually exclusive hypotheses using the law of total probability: P(B) = P(B | A)·P(A) + P(B | ¬A)·P(¬A).
Application
Identify the givens: P(cavity) = 0.2, so P(¬cavity) = 0.8; P(toothache | cavity) = 0.6; P(toothache | ¬cavity) = 0.1.
Numerator (joint): P(toothache | cavity)·P(cavity) = 0.6 × 0.2 = 0.12.
Denominator (total probability of toothache): P(toothache) = 0.6×0.2 + 0.1×0.8 = 0.12 + 0.08 = 0.20.
Apply Bayes’ rule: P(cavity | toothache) = 0.12 / 0.20 = 0.60.
Cross-check
The complementary posterior must sum to 1: P(¬cavity | toothache) = (0.1×0.8)/0.20 = 0.08/0.20 = 0.40, and 0.60 + 0.40 = 1.00. The two posteriors are consistent, confirming the result.
Answer: 0.600