Given the following probabilities:
$P(A) = 0.5$, $P(¬A) = 0.5$
$P(E|A) = 0.1$, $P(F|A) = 0.8$
$P(E|¬A) = 0.3$, $P(F|¬A) = 0.2$
I am trying to solve for $P(A|Z)$ where $Z = (E ∩ F)$. Please note that conditional independence applies here.
I've successfully found an answer by not expanding Z to it's whole form, i.e.
$P(A|Z) = \frac{P(A)P(Z|A)}{P(Z)}$ [bayes theorem]
$P(A|Z) = \frac{P(A)P(E ∩ F|A)}{P(Z)}$ [expand numerator]
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(Z)}$
Apply $P(Z) = P(A)P(Z|A)+P(¬A)P(Z|¬A)$ to denominator:
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(A)P(Z|A)+P(¬A)P(Z|¬A)}$
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(A)P(E ∩ F|A)+P(¬A)P(E ∩ F|¬A)}$ [expand denominator]
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(A)P(E|A)(F|A)+P(¬A)P(E|¬A)P(F|¬A)}$ [final expression]
Now, if I start again by expanding Z before applying $P(Z) = P(A)P(Z|A)+P(¬A)P(Z|¬A)$.
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(E ∩ F)}$
And apply conditional independence $P(E ∩ F) = P(E)*P(F|E) = P(E)P(F)$:
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{P(E)P(F)}$
and apply $P(E) = P(E|A)P(A) + P(E|¬A)P(¬A)$ (apply to $P(F)$ as well):
$P(A|Z) = \frac{P(A)P(E|A)P(F|A)}{[P(E|A)P(A) + P(E|¬A)P(¬A)][P(F|A)P(A) + P(F|¬A)P(¬A)]}$ [final expression]
Solving each of these final expressions using the given probabilities, I get two different answers:
$\frac{0.5*0.1*0.8}{0.5*0.1*0.8+0.5*0.3*0.2}$ = $P(A|E,F) = 0.57$
$\frac{0.5*0.1*0.8}{[0.1*0.5 + 0.3*0.5][0.8*0.5 + 0.2*0.5]} = P(A|E,F) = 0.40$
Any help on where I went wrong in my expansion/understanding would be greatly appreciated!
You are assuming $E$ and $F$ independent and also independent given $A.$ Then, $$ P(E) = P(E \mid A) P(A) + P(E \mid ¬A)P(¬A) = (0.1)(0.5)+(0.3)(0.5) = 0.2, $$ similarly $$ P(F) = 0.5 $$ This gives $P(EF) = (0.2)(0.5) = 0.1.$ And the assumption gives $P(EF \mid A) = (0.1)(0.8) = 0.08.$ Then $$ \begin{align*} P(EF \mid ¬A) &= 2 P(EF¬A) = 2(P(EF) - P(EFA)) \\ &= 2(P(EF)-P(E\mid A)P(F \mid A) P(A)) = 2((0.1) - (0.1)(0.8)(0.5)) \\ &= 0.12 \neq P(E \mid ¬A) P(F \mid ¬A). \end{align*} $$
So $E$ and $F$ aren't independent given $¬A,$ which is one thing you are assuming as well. I didn't check other things. At any rate, you cannot apply several steps you did apply.