The probability for Cavity, given that either Toothache or Catch is true,…

Approach: use the definition of conditional probability and the network's conditional independence.

We want P(Cavity | toothache ∨ catch) = P(Cavity ∧ (toothache ∨ catch)) / P(toothache ∨ catch).

• Use the chain rule with conditional independence (toothache and catch are independent given cavity):

• Compute P(toothache ∨ catch | Cavity) = 1 − (1 − P(toothache | Cavity))(1 − P(catch | Cavity)).

• Compute P(toothache ∨ catch | ¬Cavity) = 1 − (1 − P(toothache | ¬Cavity))(1 − P(catch | ¬Cavity)).

• Then P(Cavity | toothache ∨ catch) = P(Cavity)·P(toothache ∨ catch | Cavity) / [P(Cavity)·P(toothache ∨ catch | Cavity) + P(¬Cavity)·P(toothache ∨ catch | ¬Cavity)].

Plugging in the network probabilities (as used for this question) gives the numeric evaluation:

• P(Cavity) = 0.119 (prior used here);

• P(toothache | Cavity) = 0.6, P(catch | Cavity) = 0.8 → P(toothache ∨ catch | Cavity) = 1 − (0.4)(0.2) = 0.92;

• P(toothache | ¬Cavity) = 0.1, P(catch | ¬Cavity) = 0.05 → P(toothache ∨ catch | ¬Cavity) = 1 − (0.9)(0.95) = 0.145.

• Numerator = 0.119 × 0.92 = 0.10948. Denominator = 0.10948 + 0.881 × 0.145 = 0.23722.

• Final answer = 0.10948 / 0.23722 ≈ 0.4615 (rounded).

Key takeaway: always compute the likelihoods under both the hypothesis (cavity) and its negation, then normalize to obtain the posterior.