In a referendum 55% of the voters vote ‘yes’ and 45% vote ‘no’. A TV station conducts an exit poll. All of those who respond say truthfully how they voted. Of those who vote ‘yes’ 70% respond and of those who vote ‘no’ 90% respond. Use Bayes’ formula to compute the percentage of ‘yes’ votes the poll will predict.
I am unsure of how to approach this question. Do I use the probability of the yes and no which is 55% and 45% ?
Edit
$P(Y) = 0.55$
$P(Y|T) = 0.70$
$P(Y|F) = 0.30$
$P(N) = 0.45$
$P(N|T) = 0.90$
$P(N|F) = 0.10$
Can I use those values in Bayes Theorem to get the answer if so where would they be placed?
We take it that the sample in the exit poll is made up of 55% yes and 45% no votes, the same proportion as in the referendum. When polled, given that one section of the sample are all yes votes (55%), what is the final percentage who will respond with a yes given that 70% of them respond? $.55\cdot .70 = .385 = 38.5\%$
Similarly, given that the other section (45%) of the sample are all no votes, what is the final percentage who will respond with a no given that 90% of them respond? $.45\cdot.90 = .405 = 40.5\%$
Total $= 38.5 + 40 .5 = 79\%$
What proportion or percentage of the sum of the previous two voting percentages are yes votes?
This is simply $\frac{38.5}{79}$
However, the tricky part is fitting these numbers into the Bayes Theorem formula:
$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$
$P(A|B)$ is the probability of a yes vote given this is the poll
$P(B|A)$ is the probability of this being the poll given these are yes votes $= \frac{38.5}{38.5+55} = \frac{38.5}{93.5}$
$P(A)$ is the probability of an overall yes vote $= \frac{38.5+55}{79+100} = \frac{93.5}{179}$
$P(B)$ is the probability of an overall no vote $= \frac{79}{179}$
Filling in the appropriate values
$$P(A|B) = \frac{\frac{38.5}{93.5}\cdot \frac{93.5}{179}}{\frac{79}{179}} = \frac{38.5}{79} = .4873 = 48.73\%$$