# Given two indistinguishable urns

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 $$4/6$$?

$$P(A \cap E)$$ 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 $$P(E)$$.

• How are they independent? If E happens then A could happen too Apr 3 '19 at 15:56
• You can get a red marble out of event B. But you are correct, I should have said P(A)*P(E|A) Apr 3 '19 at 15:59
• I get that $P(A|E)=8/3$ which is impossible Apr 3 '19 at 16:07
• 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. Apr 3 '19 at 16:12
• is P(A)=1/2 and P(E|A)=1/2 ? Apr 3 '19 at 16:12

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:

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

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.