Bayes Theorem - Referendum 
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? 
 A: 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
$$P(A|B) = \frac{\frac{38.5}{93.5}\cdot \frac{93.5}{179}}{\frac{79}{179}} = \frac{38.5}{79} = .4873 = 48.73\%$$
