Experiment concerns two Urns, A and B

Urn A contains 3 Red balls and 4 White balls; Urn B contains 2 Red balls and 1 white Ball. First an urn is selected at random(with Equal Probability), and then a ball is selected at random from the urn. compute Prob (A | Red).

The answer is .39

I am getting some really off number and I cant figure this one out. I am doing a practice test for an upcoming exam any help would be great thanks guys.

You need to use Baye's theorem.

Let P(r) be the probability of getting a red ball.
P(r) = P(r|A)P(A) + P(r|B)P(B)
Now,
P(r|A) = 3/7 (there are 3 red balls in the urn out of a total of 7 balls.
P(r|B) = 2/3 (2 red balls out of 3)
We know that P(A) = (1/2) and P(B)=(1/2)

So P(r) = (3/7)(1/2) + (2/3)(1/2) = 3/14 + 1/3 = 0.547619

Bayes theorem says
P(A|r) = P(r|A)P(A) / P(r)
P(A|r) = (3/14)/0.547619 = 0.3913