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.
 A: Your first calculation for $P(H)$ is correct, and also makes sense if you think about it (I like to do this to check if an answer seems reasonable) - there are a total of 6 coin sides spread out over the 3 coins, of which 5 are heads. One is chosen completely at random with no other information - this is why the probability is $5/6$.
For the second part you have Bayes theorem correct, though since the question asks about the fair coin, you should probably have used C there. $$P(C|H)=\frac{P(H|C)P(C)}{P(H)}=\frac{\frac12\cdot\frac13}{\frac56}=\frac15$$
This again makes sense - 5 of the faces were heads, and you know only one of them was on the fair coin. All you know is you tossed a head, so the chances of that head being from the fair coin is 1 out of 5, or $1/5$. 
If instead you were after the probability of you having picked any biased coin, you'd get $4/5$, if you were after the probability of you having picked a particular biased coin, then you'd get $2/5$ as you got in your calculation.
A: you are using $C$ for the regular coin. So you should be calculating $P(C|H)$, which is
$$P(C|H)=\frac{P(H|C)*P(C)}{P(H)}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}$$
A: 
I am confused as to if P(A)⋅P(H∣A) should be 2/3 or 1/3. If P(A)⋅P(H∣A) is 2/3, then the final answer would be 4/5

If $A$ is the event "a particular one of the two biased coins" then $P(A\cap H)=\tfrac 13\tfrac 22$ as you have.  Just as $P(B\cap H)=\tfrac 13\tfrac 22$.  In this case $A\cup B$ is the event "either one of the two biased coins" and so indeed $\mathsf P((A\cup B)\cap H)~=~ \tfrac 23\tfrac 22~=~\mathsf P(A)~\mathsf P(H\mid A)+\mathsf P(B)~\mathsf P(H\mid B)$.
Thus, it is okay that: $\mathsf P(H)~{=\mathsf P(A\cup B)~\mathsf P(H\mid A\cup B)+\mathsf P(C)~\mathsf P(H\mid C) \\= \tfrac 23\tfrac 22+\tfrac 13\tfrac 12\\=\tfrac 56}$
However, this brings us to your second part. $C$ is the event of "an unbiased coin was picked", and as you have already correctly determined that $\mathsf P(C)~\mathsf P(H\mid C)=\tfrac 13\tfrac 12$ then you should have:
$$\dfrac{\mathsf P(C)~\mathsf P(H\mid C)}{\mathsf P(H)}~=~\dfrac{\tfrac 13\tfrac 12}{\tfrac 56}=\dfrac 15$$

Intuitively, each of the $5$ heads is equally likely to show up, and only $1$ of these is from the unbiased coin, so if one does show up, then there is a probability of $1/5$ for it to be from that unbiased coin.


The question is what is the probability you pick up the fair coin and it is head?

PS: That is $\mathsf P(C\cap H)=\mathsf P(C)\mathsf P(H\mid C)=\tfrac 16$
