Hello. I have been trying to calculate what the probability is that p(A|not Favor) from this problem. This is specifically referring to part b.
The results that I got were:
P(A|Favor) = 9.09%
P(B|Favor) = 22.72%
P(C|Favor) = 68.18%
So would the actual answers be that
P(A|NOT Favor) = 80.91
P(B|NOT Favor) = 77.28
P(C|NOT Favor) = 31.82
You are already given the quantities $$\Pr[A] = 0.2, \quad \Pr[B] = 0.3, \quad \Pr[C] = 0.5$$ as representing the probabilities of a randomly selected voter belonging to the respective party. You are also given $$\Pr[F \mid A] = 0.3, \quad \Pr[F \mid B] = 0.5, \quad \Pr[F \mid C] = 0.9,$$ representing the conditional probabilities of favoring ($F$) the ballot measure given membership in each respective party. The implied assumption in the question (which is not necessarily true in the real world, as voters may be indecisive) is that $$\Pr[\bar F \mid A] = 0.7, \quad \Pr[\bar F \mid B] = 0.5, \quad \Pr[\bar F \mid C] = 0.1,$$ representing the conditional probabilities of opposing ($\bar F$) the ballot measure given membership in each respective party. You are asked to compute, in parts (b) - (d), $$\Pr[A \mid \bar F], \quad \Pr[B \mid \bar F], \quad \Pr[C \mid \bar F].$$ To do so efficiently, we compute the unconditional (marginal) probability of opposing the measure; i.e., $$\Pr[\bar F] = \Pr[\bar F \mid A]\Pr[A] + \Pr[\bar F \mid B]\Pr[B] + \Pr[\bar F \mid C]\Pr[C],$$ for which you already have all of the requisite probabilities. The result should be $0.34$. Then all that remains is to compute $$\Pr[P \mid \bar F] = \frac{\Pr[\bar F \mid P]\Pr[P]}{\Pr[\bar F]}$$ where $P \in \{A, B,C\}$ for each party. It bears mentioning that your answers cannot possibly be correct, because you must have $$\Pr[A \mid \bar F] + \Pr[B \mid \bar F] + \Pr[C \mid \bar F] = 1.$$ This is because, given that a voter was selected and was found to oppose the measure, they belong to exactly one party; thus the sum of conditional probabilities must equal $1$.