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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 of the second marble being red, if the first marble we pulled is also red?

I'm not sure how to interpret this question and start my solution. Since we don't know which Urn was drawn from, does this mean that our sample space contains all $10$ marbles and the probability of the first marble being red $=$ to the probability of the second marble being red $= \frac{5}{10} = \frac{1}{2}?$ What role do the Urns play? Can someone please offer some insight? Thank you for your time.

J. Hodge
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  • Do you put the first marble back into the urn after drawing it before drawing the second one? – Alejandro Bergasa Alonso May 05 '20 at 22:43
  • @AlejandroBergasaAlonso Yes. – J. Hodge May 05 '20 at 22:44
  • Define events $R_1$ and $R_2$ as the first marble being red, and the second, respectively. Then you want $P[R_2|R_1]$. Note that both $R_1$ and $R_2$ can occur under either event $A$ (urn $A$) or event $B$ (urn $B$). To start, can you compute $P[R_1]$? Doing that will involve the probability for selecting urn A and urn B, which is what you are asking. – Michael May 05 '20 at 22:50
  • @Michael Can we define an event $C$: either urn $A$ or urn $B$ is selected with $P(C) = \frac{1}{2}$ and then calculate $P(R_1 \mid C) = \frac{P(R_1 \cap C)}{P(C)}$? Or is this also the wrong approach? – J. Hodge May 05 '20 at 23:00
  • We know that either urn A or urn B is selected. It sounds like you are defining $C=A \cup B$ and so $C$ is the entire sample space! Thus $P[C]=1$. – Michael May 05 '20 at 23:00

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Define $M_1$ and $M_2$ to be two events representing the first and the second marble being red, accordingly. We need to find the probability $P(M_2|M_1)$. Using the formula for conditional probability, we get $$P(M_2|M_1)= \frac{P(M_2\cap M_1)}{P(M_1)}.$$

Firstly consider the probability that the first marble we draw is red, i.e. denominator. It is the probability that it is red and comes from the urn A, plus the probability that it is red and comes from the urn B (this idea is called the law of total probability). More rigorously, if we define the event of drawing the marble from urn A with the set $A$ and the event of drawing the marble from urn B with the set $B$, then $A\cup B=\Omega$, $A\cap B=\emptyset$ (A and B are said to be disjoint) and $$P(M_1) = P(M_1\cap A) + P(M_1\cap B) = \frac{1}{2}\cdot\frac{3}{5} + \frac{1}{2}\cdot\frac{2}{5}= \frac{1}{2}.$$

The probability of two drawn marbles being red, i.e. numerator, is computed in the same way, $$ P(M_2\cap M_1) = P(M_2\cap M_1\cap A) + P(M_2\cap M_1\cap B) = \frac{1}{2}\cdot\left(\frac{3}{5}\right)^2 + \frac{1}{2}\cdot\left(\frac{2}{5}\right)^2 = \frac{1}{2}\cdot\frac{13}{25}.$$ Finally, $$P(M_2|M_1)=\frac{13}{25}.$$

max
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