Conditional Probability - Basic Question On question 2b. below, why does knowing the extra bit of information A increase the probability of D?


In general:
$$P(D)=P(D\mid A)P(A)+P(D\mid A^c)P(A^c)$$
Here $A^c$ denotes the event that there are no forgeries in the shipment. Under that condition it cannot be that all paintings are forgeries. So $P(D\mid A^c)=0$ and we end up with:$$P(D)=P(D\mid A)P(A)$$
This equality makes clear that $P(A)<1$ implies $P(D\mid A)>P(D)$.
