There are two urns, urn $A$ and urn $B$. In urn $A$ there are $3$ red marbles and $2$ blue ones. In urn $B$ there are $2$ red marbles and $3$ blue ones. Through a fair coin toss we select one of the urns and draw two marbles from it consecutively with replacement. We put each marble back after drawing it. What is the probability that the second marble we draw is red, if we choose urn $A$ and the first marble we draw is also red?
I defined three events:
$E_1:$ The first marble we draw is red.
$E_2:$ The second marble we draw is red.
$U_1:$ We draw from urn $A$.
$\Rightarrow P(E_2 \mid E_1 \cap U_1) = \dfrac{P((E_1 \cap U_1) \cap E_2)}{P(E_1 \cap U_1)}$.
All I know that $P(E_2) = 0.5$, How do I proceed from here? What is $P((E_1 \cap U_1) \cap E_2)$ and $P(E_1 \cap U_1)?$ How is $P(E_1 \cap U_1)$ different from $P(E_1 \mid U_1)?$ Any help would be much appreciated.
Best Answer
You are overcomplicating this. Once you choose urn A you have an urn with 3 red marbles and 2 blue ones. You are putting the first ball back in so $ P(E2∣E1 \cap U1) = P(E2∣U1) = \frac{3}{5} $.