Bayes Theorem Coin Problem There are 3 coins. One is regular (both head and tail) and the other two only have head sides. Now, flip one coin and get head. The question is what is the probability of this coin is regular one? 
Or Twisted
There are 3 coins. One is regular (both head and tail) and the other two only have head sides. Now, flip one coin and get head. The question is what is the probability you pick up the fair coin and it is head?
P(H)=P(A)⋅P(H∣A)+P(B)⋅P(H∣B)+P(C)⋅P(H∣C)
   = (1/3)(2/2) + (1/3)(2/2) + (1/3)*(1/2) = 5/6
P(A∣H)=P(A)⋅P(H∣A)/P(H)
      = (1/3)/(5/6) =2/5
I am confused as to whether this is the correct answer or not.

you are using $C$ for the regular coin. So you should be calculating $P(C|H)$, which is
$$P(C|H)=\frac{P(H|C)*P(C)}{P(H)}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}$$
