The probability that the other face is tails is? Question

You are given three coins:


*

*has heads on both faces

*has tails on both faces

*has a head on one face and a tail on the other.


You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is $?$

My Approach
This problem can be solved without using bayes' theorem ,but i want to solve it using bayes'.
let
Probability of coming head on toss=$P(H)$
Probability that other face is tail =$P(OT)$
Probability that other face is Head =$P(OH)$
So question is to find $P(\frac{OT}{H})$
$$P(\frac{OT}{H})=\frac{P(\frac{H}{OT})\times P(OT)}{P(\frac{H}{OT})\times P(OT)+P(\frac{H}{OH})\times P(OH)}$$
$P(\frac{H}{OT})=\frac{1}{2}$i.e HT among HH,HT
$P(\frac{H}{OH})=\frac{1}{2}$i.e HH among HH,HT
I am stucked what will be $P(OH)$ and $P(OT)$
please help

Label the faces of the $HH$ coin as $H_f,H_b$ where the subscripts $f,b$ stand for "front", "back", respectively.

Similarly, label the faces of the $TT$ coin as $T_f,T_b$.

For the $HT$ coin, just use the labels $H,T$.

We can regard the sample space for the experiment as the set
$$\{(H_f,H_b),\,(H_b,H_f),(H,T),(T,H),(T_f,T_b),(T_b,T_f)\}$$
where each pair represents $(\text{face}\;1,\text{face}\;2$), [face $1$ is the shown face, and face $2$ is the other face], and all $6$ ordered pairs are equally likely.

Let $H_1$ be the event of heads on face $1$, and let $T_2$ be the event of tails on face $2$.

Thus, $P(H_1|T_2)$ is what you denoted as $P(H|OT)$.
\begin{align*}
\text{Then}\;\;&P(H_1|T_2)\\[4pt]
=\;&\frac{P(H_1\cap T_2)}{P(T_2)}\\[4pt]
=\;&\frac{|H_1\cap T_2|}{|T_2|}\\[4pt]
=\;&\frac{|\{(H,T)\}|}{|\{(H,T),(T_f,T_b),(T_b,T_f)\}|}\\[4pt]
=\;&{\small{\frac{1}{3}}}\\[4pt]
\end{align*}
hence, switching back to your notation, we have $P(H|OT)={\large{\frac{1}{3}}}$, not ${\large{\frac{1}{2}}}$.

By analogous reasoning, we get $P(H|OH)={\large{\frac{2}{3}}}$.

Also, by symmetry we have $P(OT)=P(OH)={\large{\frac{1}{2}}}$.

Alternatively, by direct count of qualifying pairs in the sample space,
$$P(OT)={\small{\frac{3}{6}}}={\small{\frac{1}{2}}}$$
and similarly,
$$P(OH)={\small{\frac{3}{6}}}={\small{\frac{1}{2}}}$$

Hence, Bayes' formula yields
\begin{align*}
P(OT|H)
&=\frac
{P(H|OT)P(OT)}
{P(H|OT)P(OT)+P(H|OH)P(OH)}\\[4pt]
&=\frac
{\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)}
{\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)+\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)}
\\[4pt]
&={\small{\frac{1}{3}}}\\[4pt]
\end{align*}

Note:

As drhab points out, there's no good reason to use Bayes' formula for this problem, since the same reasoning that allowed us to get
$P(H_1|T_2)={\large{\frac{1}{3}}}$ 
could have been used, as shown below, to directly find $P(T_2|H_1)$:
\begin{align*}
&P(T_2|H_1)\\[4pt]
=\;&\frac{P(T_2\cap H_1)}{P(H_1)}\\[4pt]
=\;&\frac{|T_2\cap H_1|}{|H_1|}\\[4pt]
=\;&\frac{|\{(H,T)\}|}{|\{(H,T),(H_f,H_b),(H_b,H_f)\}|}\\[4pt]
=\;&{\small{\frac{1}{3}}}\\[4pt]
\end{align*}
