Answer
$P(B|A)=\frac{.6\times.3}{.6\times.3+.5\times .7}\approx .34$
Work Step by Step
Accordint to Bayes' theorem:
$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A')P(A')}$
Here, we have
$P(A|B)= .6$
$P(B)= .3$
$P(A|B')= .5$
Be aware, that A and B are in the definition of the theorem.
We know, that $P(B')=1-P(B)=1-.3=.7$
We can substitute into the definition:
$P(B|A)=\frac{.6\times.3}{.6\times.3+.5\times .7}\approx .34$
