0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ "How is $P(E_1\cap U_1)$ different than $P(E_1\mid U_1)$?" The one is the probability that both occurred where neither is necessarily going to occur. The other is the probability of the one occurring given that the other is guaranteed to occur. $P(E_1\cap U_1)=P(U_1)\times P(E_1\mid U_1)$ $\endgroup$
    – JMoravitz
    May 7 '20 at 12:38
  • $\begingroup$ How does this differ from the last time you asked this question? If you need further clarification, it is best to ask for it on the original post rather than making a new one. $\endgroup$
    – JMoravitz
    May 7 '20 at 12:39
1
$\begingroup$

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} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.