Stumped on a Bayes Theorem Question A certain medical syndrome is usually associated with two overlapping sets of symptoms, A and B. Suppose it is known that: P($A|B$) = 0.8 P($B|A$) = 0.9 P($B'|A'$) = 0.85 Find P($A'$|$B'$) From the algebra I have done, it seems that there is missing information. Can anyone solve this or corroborate my suspicion?

We know that $$ P(A\mid B) = \frac{P(A \cap B)}{P(B)}, $$ and vice versa, so $$ P(A \cap B) = 0.8 P(B) = 0.9 P(A). $$ On the other hand, $$ P(B' \mid A') = \frac{P(A' \cap B')}{1-P(A)} = \frac{1-P(A \cup B)}{1-P(A)}, $$ and $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ So $$ 1+P(A \cap B) -P(A)-P(B) = 0.85(1-P(A)) $$ We now have 3 equations in 3 unknowns, so they should be solvable for $P(A),P(B),P(A\cap B)$. Indeed, they solve to $$ P(A) = 2/5 \\\ P(B) = 9/20 \\\ P(A \cap B) = 9/25 $$ Now we can just use $$ P(A' \mid B') = \frac{1-P(A)}{1-P(B)}P(B' \mid A') = \frac{3/5}{11/20}\frac{17}{20} = \frac{51}{55} $$