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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.

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    Nope that's the best solution (aside from a typo where you forgot P(Urn1) in your equation). – 2017-01-23

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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.

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    Is there a way to solve this problem by avoiding the Bayes' Rule and using purely logic? – 2017-01-23
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    @verkter Well, ... you *can* argue that of the *fifty* red balls, *thirty* come from urn 1, and *because* there is no bias in selecting *between* the urns and *an equal count* of balls in each urn, then the probability of selecting *any particular* ball is identical, so the probability of having selected a red ball from urn 1 given that you selected a red ball is : $30/50$. However, I would not say that's *avoiding* using Bayes' Rule, *per se*, so much as just not explicitly calling it out. – 2017-01-23
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    @Pushkar Soni No, thank you, I already solved it with trees. – 2017-01-23
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    @Graham Kemp Got it, thank you! – 2017-01-23