Checking my work for a probability problem

Question

so I was posed the question:

I have two coins. One is a fair coin, equally likely to come up heads or tails; the other is a two-headed coin. Without telling you whether it is the fair or unfair coin, a take a particular coin C and flip it twice. Both times it comes up heads. What is the probability that I have flipped the fair coin?

I worked through the problem (shown below) but I stink at math, so I wanted to get a better set of brains to see if I am way off.

To set up, H=chance of getting heads A=chance of getting 2-headed coin B=chance of getting fair coin

P(A∣H)=P(A) * P(H∣A) / P(H) --> Bayes Theorem to solve for the 2-headed then squaring (for the multiple flips) and subtracting from 1 to get the chance of a fair coin.

P(A) = 1/2, because 1 out of 2 coins are 2-headed

P(H|A) = 1, because given you have the 2-headed coin, then it is for sure (100%) you will get a head

P(H) = 3/4, because there are 3 heads and 1 tails in total, so it's 75% chance to get heads

Therefore, P(A|H) = (1/2) * 1 / (3/4) = 4/9

Subtract this from 1 to get 5/9 and square to account for the two coin flips to get 0.3086 or about a 31% chance of it being a fair coin.

Is any of this even remotely close to being correct?

Answer 1

There are a total of eight possibilities, with * denoting our desired outcome:

1: You picked the fair coin
    1a: First flip is a H
       *1ai: Second flip is a H
        1aii: Second flip is a T
    1b: First flip is a T
        1bi: Second flip is a H
        1bii: Second flip is a T
2: You picked the 2H coin
    2a: First flip is a H
       *2ai: Second flip is a H
       *2aii: Second flip is a H
    2b: First flip is a H
       *2bi: Second flip is a H
       *2bii: Second flip is a H

As we can see, there are a total of 5 different ways to get the outcome we desire, only 1 of which resulted from picking the fair coin. Therefore the answer is $\frac{1}{5}$.