I'm trying to solve problem 16 (Ch. 5) of Jeffrey's Logic of Decision. The problem says:
Suppose that A and B are pairwise incompatible propositions, and suppose that the preference ranking is as follows:
Suppose that des G = 1 and des T = 0
b) Assuming that des ¬G = -2, find des ¬A, prob A, prob B, prob G.
My attempt to solve it:
By (5-5) prob G = 1 / (1 - (des G / des ¬G))
So prob G = 1 / (1 - (1 / -2)) = 2/3
By problem 11, if A and B are ranked together but not with T, then prob A = prob B iff ¬A and ¬B are ranked together, which is the case.
By (5-6) des ¬(A∨B) = -(prob A∨B / prob ¬(A∨B)) * des A∨B = -2, since it's ranked together with ¬G.
By (5-2) des A∨B = (prob A * des A + prob B * des B) / (prob A + prob B) = 1, since prob A = prob B, and both des A and des B are equal to 1.
So, by substitution:
-2 = -(prob A∨B / prob ¬(A∨B)) * (prob A * des A + prob B * des B) / (prob A + prob B)
= -(prob A∨B / prob ¬(A∨B)) * 1.
That's as far as I get.
I've also tried to solve for ¬(A∨B), since it is logically equivalent to ¬A¬B, with equation (5-1)(e) page 81 but it doesn't get me anywhere.
What am I missing?
Would appreciate any help.
Thanks for your time and consideration.
NOTE: It's not homework. It's just me trying to understand the theory.