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 |
- A.
0.400
- B.
0.600
- C.
0.280
- D.
0.216
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Show answer & explanation
Correct answer: B
For events A and B with P(B) > 0, the conditional probability is defined as P(A | B) = P(A ∧ B) / P(B). When a full joint probability distribution is available, P(B) is found by summing every joint-probability cell in which B holds, and P(A ∧ B) is found by summing only the cells in which both A and B hold together.
Apply this to P(toothache | cavity) using the joint distribution above:
Read off the cavity row of the table: the four joint probabilities are 0.108 (toothache, catch), 0.012 (toothache, ¬catch), 0.072 (¬toothache, catch), and 0.008 (¬toothache, ¬catch).
Sum the whole row to get P(cavity): 0.108 + 0.012 + 0.072 + 0.008 = 0.200.
Sum only the toothache-present cells within that row to get P(toothache ∧ cavity): 0.108 + 0.012 = 0.120.
Divide: P(toothache | cavity) = P(toothache ∧ cavity) / P(cavity) = 0.120 / 0.200 = 0.600.
Cross-check: the remaining cavity-row cells give the complementary probability, (0.072 + 0.008) / 0.200 = 0.400. Since toothache and ¬toothache are the only two possibilities given cavity, the two conditional probabilities must add to 1 — and 0.600 + 0.400 = 1.000, confirming the answer.