Why can't I apply the odds version of Bayes' to this exit poll problem?

Question

This is a question about an alternative solution to the problem stated in Probability and exit polls

The problem is:

Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let A be the event that they voted for A, and W the event that they are willing to participate in the exit poll. Suppose that P(W∣A)=0,7 but P(W∣AC)=0,3. In the exit poll, 60% of the respondents say they voted for A (assuming they are all honest), suggesting a comfortable victory for A. Find P(A).

The answer there uses Law of Total probability to arrive at an answer of 9/23. However, I'm confused why when I use the odds form of Bayse' rule, I get a completely different answer.

This is what I do:

$P({W}\mid{A})=0.7$ (probability of participating in exit poll given voting for A)

$P({W}\mid{A^c})=0.3$ (probability of participating in exit poll given voting for B)

$\frac{P({A}\mid{W})}{P({A^c}\mid{W})}=0.6$ (portion of respondents that say they voted for A in the exit poll)

I'm looking for $P(A)$. So, applying the odds form of Bayse' rule:

$\frac{P({A}\mid{W})}{P({A^c}\mid{W})}=\frac{P({W}\mid{A})}{P({W}\mid{A^c})} * \frac{P(A)}{P(A^c)}$

$\frac{6}{10} = \frac{7}{3} * \frac{P(A)}{1-P(A)}$

$\frac{9}{35} = \frac{P(A)}{1-P(A)}$

$P(A)\approx0.2$

That's far off from $\frac{9}{23}$. I'm wondering where I'm going wrong?

Answer 1

$\frac{P({A}\mid{W})}{P({A^c}\mid{W})}=\frac{P({W}\mid{A})}{P({W}\mid{A^c})} * \frac{P(A)}{P(A^c)}$

Here $P({A}\mid{W}) = 0.6$ and $P({A^c}\mid{W}) = 0.4$

so equation should be $\frac{0.6}{0.4} = \frac{7}{3} \frac{x}{1-x}$ where $x = P(A)$. This gives $x = \frac{9}{23}$.