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2015-05-15

Practice exam #6.

We know that P(Cavity | toothache) = α[P(Cavity, toothache)].

Lets say P1 = P(cavity ∧ toothache), P2 = P(¬cavity ∧ toothache). We already know both probabilities since we know P(Cavity, toothache).

So we have P(Cavity | toothache) = α[<P1, P2>]. We just need to solve for α now.

α(P1 + P2) = 1

α = 1 / (P1+ P2)

We know that P(Cavity | toothache) = α[P(Cavity, toothache)]. Lets say P1 = P(cavity ∧ toothache), P2 = P(¬cavity ∧ toothache). We already know both probabilities since we know P(Cavity, toothache). So we have P(Cavity | toothache) = α[<P1, P2>]. We just need to solve for α now. α(P1 + P2) = 1 α = 1 / (P1+ P2)

2015-05-18

-- Practice exam #6

Aaron Shaw, Terence Ramos, Shinjo Melosh were also in my group.