Confusion on the conditional probability- regular vs two-headed coin I am doing a question on probability (Example 1.27) from this book.  I was trying the approach suggested by the book to solve the question.
The question:

A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). I choose a coin at random and toss it twice. Define the following events:
A= First coin toss results in an H.
B= Second coin toss results in an H.
C= Coin 1 (regular) has been selected.
Find P(A|C).

My approach is first writing down the sample space $S = \{(H_R,H_R), (T_R,H_R),(H_R,T_R),(T_R,T_R),(H_F,H_F)\} $ and $A = \{(H_R,H_R),(H_R,T_R),(H_F,H_F)\} $, $B=\{(H_R,H_R), (T_R,H_R),(H_F,H_F)\} $ and $C= \{R,F\} $
According to the book, $P(A \cap C) = \frac{|A\cap C|}{|S|} = \frac{2}{5}  $ and $P(C) = \frac{|C|}{|S|}= \frac{4}{5}$. So $P(A|C) = \frac{1}{2}$, which makes sense until that point.
However, what confuses me is that the book later gives $P(C) = \frac{1}{2}$, which is totally different compared to the sample space approach. To me, that makes sense too because there are only two coins to choose from. But why do the two different approaches give two different answers?
Am I missing something?
 A: 
My approach is first writing down the sample space

Don't do that.  The outcomes you chose do not have equal probability weights.  You can not just count them.
The outcome of $(H_F, H_F)$ is four times as "heavy" as each of the other four outcomes.
$\mathsf P(A\mid C)= 1/2$ because when you have a regular coin, the probability that the first toss shows heads is $1/2$.
$$\small\mathsf P(A\mid C)=\dfrac{\lvert\{(H_R,H_R),(H_R,T_R)\}\rvert}{\lvert\{(H_R,H_R),(H_R,T_R),(T_R,H_R),(T_R,T_R)\}\rvert}=\dfrac{2}{4}$$
$\mathsf P(A\mid C^\complement)=1$ because when you have a false coin, it will always show heads.
$$\small\mathsf P(A\mid C^\complement)=\dfrac{\lvert\{(H_F,H_F)\}\rvert\times 4}{\lvert\{(H_R,H_R)\}\rvert\times 4}=\dfrac{4}{4}$$
$\mathsf P(C)=1/2$ because when you select from the two coins without bias, the probability for selecting the regular coin is $1/2$.
$$\small\mathsf P(A\mid C)=\dfrac{\lvert\{(H_R,H_R),(H_R,T_R),(T_R,H_R),(T_R,T_R)\}\rvert}{\lvert\{(H_R,H_R),(H_R,T_R),(T_R,H_R),(T_R,T_R)\}\rvert+\lvert\{(H_R,H_R)\}\rvert\times 4}=\dfrac{4}{8}$$
