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

 A: Because now the sample space is reduced. The sample space now no longer contains the event that all 5 paintings are authentic. Considering that the probability of none being a forgery is as large as 0.76, and as this event is ruled out, there would be a definite change in the other probabilities.
A: From the question I get the feeling that you want an answer regarding intuition and not so much mathematics. If I'm wrong, please comment and I will change my answer. 
From the probabilities given, we see that it is much more likely that the whole shipment consists of forgeries (10%) than it is likely that only 2,3 or 4 paintings are forgeries (2%, 1% and 2%). Hence, knowing that there is at least 1 forgery in the shipment, increases the chance that the whole shipment is "tainted".
A: 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 $P(D\mid A)>P(D)$.
