nP(Toothache, Cavity, Catch) has 23 – 1 = 7 independent entries
n
nIf I have a cavity, the probability that the probe
catches in it doesn't depend on
whether I have a toothache:
n(1) P(catch | toothache, cavity) = P(catch | cavity)
n
nThe same independence holds if I haven't got a
cavity:
n(2) P(catch | toothache, Øcavity) = P(catch | Øcavity)
n
nCatch is conditionally independent of Toothache given Cavity:
nP(Catch | Toothache,
Cavity) = P(Catch | Cavity)
n
nEquivalent statements:
nP(Toothache | Catch, Cavity) = P(Toothache | Cavity)
nP(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)