Let's say I have two urns. Urn #1 contains 30 red marbles and 10 black marbles. Urn #2 contains 20 red marbles and 20 black marbles. I choose an urn at random, then choose a random marble from the urn. If the marble I chose is red, what is the probability that I chose Urn #1?
My approach: $$P(Urn 1|Red) = \frac{P(Urn1)P(Red| Urn 1)}{P(Red)} = \frac{\frac{1}{2}*\frac{3}{4}}{(\frac{1}{2}*\frac{3}{4})+(\frac{1}{2}*\frac{1}{2})}=\frac{3}{5}$$
Is there a more practical logical and/or formula approaches to tackle this problem?
Edit: Added P(Urn1) to fix the formula.
Indeed, aside from the missing term in the numerator your solution is correct (so it is likely you knew it should be there and just missed typing it).
To be sure, your solution should read something like: $$\begin{align}\mathsf P(\operatorname{Urn}_1\mid\operatorname{Red})~&=~\dfrac{\mathsf P(\operatorname{Urn_1})\cdotp\mathsf P(\operatorname{Red}\mid\operatorname{Urn_1})}{\mathsf P(\operatorname{Red})} && \text{Bayes' Rule}\\[1ex]&=~\dfrac{\mathsf P(\operatorname{Urn_1})\cdotp\mathsf P(\operatorname{Red}\mid\operatorname{Urn_1})}{\mathsf P(\operatorname{Urn_1})\cdotp\mathsf P(\operatorname{Red}\mid\operatorname{Urn_1})+\mathsf P(\operatorname{Urn_2})\cdotp\mathsf P(\operatorname{Red}\mid\operatorname{Urn_2})} && \text{Law of Total Probability}\\[1ex] &=~\dfrac{\tfrac 12\cdotp\tfrac {30}{40}}{\tfrac 12\cdotp\tfrac {30}{40}+\tfrac 12\cdotp\tfrac {20}{40}} && \text{From the Data}\\[1ex]&=~\dfrac 35\end{align}$$ That is all.