Probability and exit polls

Question

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?

Answer 1

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!

Answer 2

@Soren123 Adding to Tim's answer and to your query that we don't have any information about P(A|W^c), 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(A^c)=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|A^c) 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*}