The probability of a toothache, given evidence of a cavity, P(toothache |…
2021
The probability of a toothache, given evidence of a cavity, P(toothache | cavity), is __________.
The full joint probability distribution for the Boolean variables Cavity, Toothache, and Catch is given below:
toothache ∧ catch | toothache ∧ ¬catch | ¬toothache ∧ catch | ¬toothache ∧ ¬catch | |
|---|---|---|---|---|
cavity | 0.108 | 0.012 | 0.072 | 0.008 |
¬cavity | 0.016 | 0.064 | 0.144 | 0.576 |
Answer: B. 0.600 — ConceptFor events A and B with P(B) > 0, conditional probability is P(A | B) = P(A ∧ B) / P(B). In a full joint distribution, the denominator is the sum of…
- A.
0.400
- B.
0.600
- C.
0.280
- D.
0.216
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Show answer & explanation
Correct answer: B
Concept
For events A and B with P(B) > 0, conditional probability is P(A | B) = P(A ∧ B) / P(B). In a full joint distribution, the denominator is the sum of every cell where B holds, while the numerator is the sum of only those cells where A and B hold together.
Application
Restrict the table to the cavity row: 0.108, 0.012, 0.072, and 0.008.
Add all four entries to obtain P(cavity) = 0.108 + 0.012 + 0.072 + 0.008 = 0.200.
Within that row, add the two entries where toothache is present: P(toothache ∧ cavity) = 0.108 + 0.012 = 0.120.
Apply the definition: P(toothache | cavity) = 0.120 / 0.200 = 0.600.
Cross-check
The complementary cavity-row probability is (0.072 + 0.008) / 0.200 = 0.400. Because toothache and ¬toothache exhaust the possibilities once cavity is given, 0.600 + 0.400 = 1.000, confirming the result 0.600.