Bag $1$ has $3$ black and $5$ white balls. Bag $2$ has $6$ black and $4$ white balls. A bag is selected at random and two balls are found to be black. Find the probability that bag $1$ was selected.
Attempt: $$\frac{\frac{3}{8} \cdot \frac{3}{8}}{\frac{3}{8} \cdot \frac{3}{8} + \frac{6}{10} \cdot \frac{6}{10}}$$
On
If we look at probability of bags producing 2 black balls individuali bag 1 would have 3/8*2/7=3/28, while bag 2 would have 6/10*5/9=1/3. Since we focus on just a situation where 2 black balls are drawn, we can ignore all other situations. Adding two probabilities we get 37/84. We can now use that to divide probability of draving 2 black balls from bag 1 and get around 24.3243%
We want to find the probability that the two black balls were extracted from bag 1 given that two black balls were extracted from a randomly chosen bag. Let $E$ be the event that two black balls were extracted and $F$ be the event that the balls were extracted from bag 1. \begin{align*} \Pr(F \mid E) & = \frac{\Pr(E \cap F)}{\Pr(E \cap F) + \Pr(E \cap F')}\\ & = \frac{\Pr(F)\Pr(E \mid F)}{\Pr(F)\Pr(E \mid F) + \Pr(F')\Pr(E \mid F')}\\ & = \frac{\frac{1}{2} \cdot \frac{\binom{3}{2}}{\binom{8}{2}}}{\frac{1}{2} \cdot \frac{\binom{3}{2}}{\binom{8}{2}} + \frac{1}{2} \cdot \frac{\binom{6}{2}}{\binom{10}{2}}} \end{align*} where the calculations are based on the assumption that the balls are selected without replacement.
If selection with replacement was intended, the author should have explicitly stated that. Under that interpretation, your answer would have been correct.