I am just preparing for some interviews and came across this questions:
There are 6 X chips in Urn A and 6 Y Chips. 3 X chips in Urn B and 9 Y chips. Given that the chip picked is X, what is the probability of picking a X chip again and what is the estimated value of this pick. (You do not replace the chip, so once it is has been picked you decrease the number of chips in the relevant Urn.) Additional assumptions: The first chips is selected with uniform probability and the second pick can be from any of the Urns.
I have so far concluded that I will be using Bayes theorem to solve it and this is how far I got:
Let A represent Urn A and B represent Urn B. $P(X|X) = ? $
$P(X) = P(A) \times P(X|A) + P(B) \times P(X|B)= \frac{3}{8}$
$P(A|X) = \frac{P(X|A)\times P(X)}{P(A)} = \frac{3}{8}$
$P(B|X) = \frac{P(X|B)\times P(X)}{P(B)} = \frac{3}{16}$
$P(X|X) = P(A|X) \times P(X) + P(B|X) \times P(X) = \frac{15}{88} + \frac{6}{176} $
I am not sure if my logic is correct, can anyone please help out with this and the second part of the question?
If the second chip is drawn from either urn with uniform probability across chips, then whether the first chip was drawn from urn A or B is irrelevant and the probability of the second draw being a chip of type X is simply $8/24 = 1/3$. If the second chip is drawn from either urn with uniform probability across urns, then chips within urns--that is to say, you first choose the urn to draw from with equal probability of choosing urn A versus urn B, then choose within that urn a chip with equal probability--then the probability of the second draw being a chip of type X will depend on the urn from which the first chip was drawn. But it was not explicitly stated in your question that this was the structure of the draws.
In this second, more mathematically interesting scenario, we informally observe that, given the first draw was a chip of type X, it is twice as likely that urn A now contains 5 X and 6 Y chips as it is that urn B now contains 2 X and 9 Y chips. That is to say, there are two possible states of the urns after the first draw, and the first state $U_1$ has probability $2/3$ of occurring, and the second $U_2$ has probability $1/3$. Consider the first state: we then have that the probability of drawing a second chip of type $X$ is $$\Pr[X^* \mid U_1] = \frac{5}{11}\cdot\frac{1}{2} + \frac{3}{12}\cdot\frac{1}{2} = \frac{31}{88}.$$ In the second state, $$\Pr[X^* \mid U_2] = \frac{6}{12}\cdot\frac{1}{2} + \frac{2}{11}\cdot\frac{1}{2} = \frac{15}{44}.$$ Therefore, $$\Pr[X^* \mid X] = \frac{31}{88}\cdot\frac{2}{3} + \frac{15}{44}\cdot\frac{1}{3} = \frac{23}{66}.$$