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Given two indistinguishable urns. The first contains a white marble and a red marble and the other urn contains three red ones and a green one. A urn is randomly chosen and one marble is extracted. Knowing that the marble extracted is red, calculate:

(a) the probability that the first urn was chosen;

(b) the probability that the second urn has been chosen.

So saying that:

$E=${the extracted marble is red}

$A=${the first urn was chosen}

$B=${the second urn was chosen}

I have to calculate (a) $P(A|E)$ which is equal to $P(A∩E)/P(E)$. I don't know how to calculate $P(A∩E)$, is it equal to $1/2$? Also, what is $P(E)$ equal to? Is it equal to $4/6$?

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$P(A \cap E)$ is $P(A)*P(E|A)$. For $P(E)$ find $P(E \cap A)$ and $P(E \cap B)$. These events are mutually exclusive, so you can just add them to find $P(E)$.

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  • $\begingroup$ How are they independent? If E happens then A could happen too $\endgroup$
    – user658422
    Apr 3 '19 at 15:56
  • $\begingroup$ You can get a red marble out of event B. But you are correct, I should have said P(A)*P(E|A) $\endgroup$
    – RandyF
    Apr 3 '19 at 15:59
  • $\begingroup$ I get that $P(A|E)=8/3$ which is impossible $\endgroup$
    – user658422
    Apr 3 '19 at 16:07
  • $\begingroup$ Then try again. The final step should be of the form x/(x+y) with x, y > 0, so I don't buy that it's greater than 1. $\endgroup$
    – RandyF
    Apr 3 '19 at 16:12
  • $\begingroup$ is P(A)=1/2 and P(E|A)=1/2 ? $\endgroup$
    – user658422
    Apr 3 '19 at 16:12
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Use Bayes' rule:

$$P(A|E) = \frac{P(E|A) P_0(A)}{P(E)} $$

Where $P_0(A)$ is the a priori probability of choosing the first urn and it's obviously $\frac{1}{2}$, $P(E|A)$ is the probability of extracting red given that the first urn is selected ($\frac{1}{2}$ again).

$P(E)$ is the a priori probability of extracting red and can be calculated as:

$$P(E) = P(E|A) P_0(A) + P(E|B) P_0(B)$$

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I think you can break it up into parts.

Let A = first urn, B = second urn.

P(A)=P(B)=1/2.

Given A, chance of getting a white ball is P(W|A)=1/2.

P(W)=P(W|A)P(A)=1/4. There's .5 probability of selecting the urn and .5 of picking white.

P(R)=P(R|A)P(A)=1/4

P(G)=P(G|B)P(B)=(1/4)(1/2)=1/8.

P(B)=P($R_3$|B)P(B)+P(G).

So P($R_3$)=P($R_3$|B)P(B)=3/8

If we pull a Red ball then a green or white ball was nut pulled.

There's an equal chance of having pulled 1 red ball from A or 1 of the 3 red balls in B.

We don't know which ball it is, but there is just a 1/4 chance it is from Urn A. That leaves a 3/4 chance it is from Urn B.

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