Use the following full joint distribution for three Boolean variables: Cavity,…
2021
Use the following full joint distribution for three Boolean variables: Cavity, Toothache, and Catch.
Cavity | Toothache | Catch | Probability |
|---|---|---|---|
Cavity | Toothache | Catch | 0.108 |
Cavity | Toothache | ¬Catch | 0.012 |
Cavity | ¬Toothache | Catch | 0.072 |
Cavity | ¬Toothache | ¬Catch | 0.008 |
¬Cavity | Toothache | Catch | 0.016 |
¬Cavity | Toothache | ¬Catch | 0.064 |
¬Cavity | ¬Toothache | Catch | 0.144 |
¬Cavity | ¬Toothache | ¬Catch | 0.576 |
The probability of Cavity, given that either Toothache or Catch is true, P(Cavity | Toothache ∨ Catch), is ______.
- A.
0.6000
- B.
0.5384
- C.
0.8000
- D.
0.4615
Attempted by 15 students.
Show answer & explanation
Correct answer: D
Concept: For events A and B with P(B) > 0, conditional probability is P(A | B) = P(A ∩ B) / P(B).
When a full joint distribution is given, obtain each required probability by summing the mutually exclusive rows that satisfy the event conditions.
Application: Let E = Toothache ∨ Catch.
For Cavity ∩ E, the qualifying Cavity rows are (Toothache, Catch), (Toothache, ¬Catch), and (¬Toothache, Catch). Therefore, P(Cavity ∩ E) = 0.108 + 0.012 + 0.072 = 0.192.
For E, include every row except those with both ¬Toothache and ¬Catch. Thus, P(E) = 0.108 + 0.012 + 0.072 + 0.016 + 0.064 + 0.144 = 0.416.
Apply the conditional-probability formula: P(Cavity | E) = 0.192 / 0.416 = 6/13 ≈ 0.4615.
Cross-check: The complementary mass within E is 0.016 + 0.064 + 0.144 = 0.224, so P(¬Cavity | E) = 0.224 / 0.416 = 7/13. The two conditional probabilities sum to 6/13 + 7/13 = 1.
Result: P(Cavity | Toothache ∨ Catch) ≈ 0.4615.