problem 34

(a): $A$ and $B$ are not independent, since knowing that $A$ occured makes $G^{c}$ more likely, which in turn makes $B$ makes more likely.
(b): $P (G ∣ ∣ A^{c}) = \frac{P (G) P (A^{c} | G)}{P (G) P (A^{c} | G) + P (G^{c}) P (A^{c} | G^{c})} = \frac{g (1 - p_{1})}{g (1 - p_{1}) + (1 - g) (1 - p_{2})}$
(c): $P (B ∣ ∣ A^{c}) = P (G ∣ ∣ A^{c}) P (B ∣ ∣ G, A^{c}) + P (G^{c} ∣ ∣ A^{c}) P (B ∣ ∣ G^{c}, A^{c}) = \frac{g (1 - p_{1})}{g (1 - p_{1}) + (1 - g) (1 - p_{2})} p_{1} + (1 - \frac{g (1 - p_{1})}{g (1 - p_{1}) + (1 - g) (1 - p_{2})}) p_{2}$