Two identical boxes contain biased coins. Each coin in box $1$ shows head with probability $1/4$, and tails with probability $3/4$, while each coin in box $2$ shows heads with probability $1/3$, and tails with probability $2/3$. Twice we choose one box at random and take a coin from that box. Each of the two coins is then tossed once. If the two coins show the same side, what is the probability that the two coins were taken from the same box?
Currently, I have the following.
Let
$A = \{\text{Box 1 is chosen}\}$, $B = \{\text{Box 2 is chosen}\}$, $C = \{\text{A coin shows heads}\}$, $D = \{\text{A coins shows tails}\}$, $E = \{\text{Both coins show the same side}\}$, $F = \{\text{Both coins are taken from the same box}\}$
Then we must determine the following. $P(F|E) = \frac{P(E|F)P(F)}{P(E)}$
Did I solve the following correctly?
$P(F) = 1/2$
$P(C)=P(C|A)P(A)+P(C|B)P(B) = 1/8 + 1/6 = 7/24$
$P(D)=P(D|A)P(A)+P(D|B)P(B) = 3/8 + 1/3 = 17/24$
$P(E)=P(C)P(C) + P(D)P(D) = \frac{7}{24}\frac{7}{24}+\frac{17}{24}\frac{17}{24}=169/288$
I am having trouble solving the following.
$P(E|F)=P(C|A)P(C|A)+P(D|A)P(D|A)+P(C|B)P(C|B)+P(D|B)P(D|B)=1/16+9/16+1/9+4/9=85/72$
What am I doing wrong?