I also found the second part of exercise 2.3 surprisingly difficult, it took much longer than I would have expected (especially now that I've figured it out and see how simple the solution is).

Exercise 2.3. Limits on Probability Values. As soon as we have the numerical values a = P(A|C) and b = P(B|C), the product and sum rules place some limits on the possible numerical values for their conjunction and disjunction. Supposing that a≤b, show that the probability of the conjunction cannot exceed that of the least probable proposition: 0≤P(AB|C)≤a, and the probability of the disjunction cannot be less than that of the most probable proposition: b≤P (A + B|C)≤1. Then show that, if a + b > 1, there is a stronger inequality for the conjunction; and if a+b < 1 there is a stronger one for the disjunction.

I'll post my solutions for the second half (proving the stronger inequalities for the conjunction and disjunction).

1.) if a+b>1, we can prove a tighter lower bound on P(AB|C) as follows:

By the generalized sum rule (2-48), we have

P(AB|C)=a+b-P(A+B|C)

and since P(x)≤1 for any x, we have

a+b-1≤a+b-P(A+B|C)=P(AB|C)

and since 1<a+b, we know 0<a+b-1, so this is a tighter bound.

The intuitive explanation for this inequality is that if a+b>1, there must be some minimum amount of "overlap" of states which satisfy both A and B. This minimum is given by a-P(notB), because the largest possible proportion of states which can satisfy A without satisfying B is P(notB), so their difference is the minimum overlap.

2.) If a+b<1, we can prove a tighter upper bound on P(A+B|C) as follows:

This is basically the same as the previous one, I'll just one-line it:

P(A+B|C)=a+b-P(AB)≤a+b<1

(yes, I realize this post is more than a year old. Hopefully others working through the book later will find my response useful)

OK, in that case I'll wait a bit to give folks time to gnaw on this bone.