Probability and exit polls I have a very simple probability question that I for some reason just can not solve.
Question: 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 $P(W \mid A)=0,7$ but $P(W \mid A^C)=0,3$. 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 $P(A)$.
Okay first we notice that $A,A^c$ obviously is a partition so we can use the total low of probability. getting
\begin{align*}
P(W) & = P(W\mid A) \cdot P(A)+P(W \mid A^c) \cdot P(A^c)\\
     & = P(W\mid A) \cdot P(A)+P(W \mid A^c)\cdot (1-P(A))\\
     & =0,7 \cdot P(A)+0,3\cdot (1-P(A))
\end{align*}
Thus all I need is to find $P(W)$ and then solve for $P(A)$. However I have issue with finding $P(W)$. Any help?
 A: You are on the right track with your use of the total law of probability.
Your equation is $P(W)=0.7P(A)+0.3[1-P(A)]$, which simplifies to $P(W)=0.4P(A)+0.3$.
The key is realizing what the "60% of respondents" information tells us. This describes another conditional probability.
60% of the respondents (those willing to participate in the exit poll) say they voted for A, so $P(A\mid W)=0.6$ and $P(A^{C}\mid W)=0.4$.
Edit: What an embarrassing mistake! I'll leave the error below, since that approach, although wrong, will make another good question...
We seek $P(A)$, so let $P(A)=x$. Then $P(A^{C})=1-x$.
From the definition of conditional probability, we know that 
\begin{align*}
P(A\mid W)&=\frac{P(A\cap W)}{P(W)}\\
&=\frac{P(A)\cdot P(W\mid A)}{P(W)}\\
&=\frac{0.7x}{0.4x+0.3}\ ,
\end{align*}
and this expression must equal 0.6. Solving the resulting equation:
\begin{align*}
\frac{0.7x}{0.4x+0.3}=0.6\\
0.7x&=0.24x+0.18\\
0.46x=0.18\\
x=9/23
\end{align*}
Thus, $P(A)=9/23$. Candidate A has less than 40% of the vote, yet the exit polling skews in A's favor. A warning for exit poll design!
End edit
Incorrect approach: Using the law of total probability, we have 
\begin{align*}
P(A)&=P(A\mid W)\cdot P(W)+P(A\mid W^{C})\cdot P(W^{C})\\
&=0.6P(W)+0.4[1-P(W)]\\
&=0.2P(W)+0.4\ .
\end{align*}
We now solve the system of equations by substituting $P(W)=0.4P(A)+0.3$ into $P(A)=0.2P(W)+0.4$.
Thus,
\begin{align*}
P(A)&=0.2P(W)+0.4\\
P(A)&=0.2[0.4P(A)+0.3]+0.4\\
P(A)&=0.08P(A)+0.06+0.4\\
0.92P(A)&=0.46\\
P(A)&=0.5\ .
\end{align*}
So the results of the exit poll are skewed by the willingness of the participants. In reality, Candidates A and B each are receiving 50% of the votes.
What a good challenging problem!
A: @Soren123 Adding to Tim's answer and to your query that we don't have any information about P(A|Wc), We can use "Law of Total Probability" to condition on A with given information, W then:
\begin{align*}
P(A)&=P(A\mid W)\cdot P(W)+P(A\mid W^{C})\cdot P(W^{C})\\
&P(A)=0.6.P(W)+P(A\mid W^{C})\cdot P(W^{C})\\
\end{align*}
Now, Lets assume P(A) = x then P(Ac)=1-x then:
\begin{align*}
P(A\mid W^{C})&=\frac{P(A)\cdot P(W^{C}\mid A)}{P(W^{C})}\\
&=\frac{(1-0.7).x}{(1-0.7).x+(1-0.3).(1-x)}\\
&=\frac{0.3.x}{P(W^{C})}\
\end{align*}
Note: We don't calculate P(W|Ac) value as it will cancel out in next step
\begin{align*}
P(A)&=P(A\mid W)\cdot P(W)+P(A\mid W^{C})\cdot P(W^{C})\\
&x=0.6.[0.4x + 0.3]+\frac{0.3.x}{P(W^{C})}.P( W^{C})\\
&x=0.24x + 0.18+0.3x\\
&0.46x=0.18\\
&P(A) = x =\frac{0.18}{0.46} = \frac{9}{23}\\
\end{align*}
