A town is composed of $2/5$ out of town couples and $3/5$ in town couples. If a couple is from out of town, the probability that the husband and wife will give you the correct directions independently is $3/4$. If the couple is from in town, they will always give you the correct direction independently of each other. Suppose you ask a random couple for directions. If both the husband and wife give you the same direction, what is the probability that this direction is correct?
I've done $X$ = (probability that direction is correct)/(probability they give you the same direction).
(probability that direction is correct) = $3/5 + (2/5)(3/4)(3/4) = 33/40$
(probability they give you the same direction) = $3/5 + (2/5)(3/4)(3/4)+(2/5)(1/4)(1/4) = 17/20$
$X = (33/40)/(17/20) = 33/34$
Does this seem correct? Have I done anything incorrectly?
Your answer is right, but consider to be more precise and use the mathematical expressions and formulas:
What you are looking for is the conditional probability $P (A\mid B) $ for $$A=\text { The couple tells you the right direction.}$$ and $$B=\text { The husband and the wife are telling the same direction.}$$
Since $$ P (A\mid B)=\frac{P (A\cap B)}{P (B)}$$ you need at first the probability that the husband and the wife are telling you the same and that this is the right direction. Therefore $$ P (A\cap B)=\frac{3}{5}+ \frac{2}{5}\cdot \frac {3}{4} \cdot\frac{3}{4}= \frac {33}{40} $$ Secondly, you need $$ P (B)= \frac{3}{5}+ \frac{2}{5}\cdot \frac {3}{4} \cdot\frac{3}{4}+ \frac{2}{5}\cdot \frac {1}{4}\cdot \frac{1}{4} =\frac{34}{40}$$ Finally $$ P (A\mid B)= \frac{33}{34}$$