I know this question has been asked, but I did not find the answer satisfactory.
An urn contains 5 white and 4 black balls. 4 balls are drawn from this urn and put into another urn. From this second urn a ball is drawn and is found to be black and it is not replaced. What's the probability of drawing a white ball at the next draw.
If we call our events A: get a black on the first draw and B: get a white on the second:
$$P(B|A) = \sum_{i=1}^4P(B|E_i)P(E_i)$$ where $E_i$ is the probability that i black balls were put into urn 2 originally.
$$P(B|A) = \sum_{i=1}^4P(B|E_i)P(E_i) = \sum_{i=1}^3P(B|E_i)P(E_i) = P(B|E_1)P(E_1)+P(B|E_2)P(E_2)+P(B|E_3)P(E_3)$$
this is equal to: $${{3 \choose1} \over {3 \choose 1}}{{5 \choose4} \over {9 \choose 4}} + {{4 \choose2}{2 \choose1} \over {3 \choose 1}}{{5 \choose2} \over {9 \choose 4}} + {{1 \choose1}{4\choose3} \over {3 \choose 1}}{{5 \choose1} \over {9 \choose 4}} = ...$$
Is this the correct way to approach this question? Thanks for your time.
As noted in the linked answer, the second urn is not needed to solve the problem, in which case the answer is obviously 5/8. I wanted to arrive at the same answer via conditional probabilities (since the problem was in a chapter on the same subject). This is the solution:
Let A: event that the second is white, B: Event that the first is black, $E_i$: the event that i white balls were transferred.
$P(E_0) = {1\over126}$, $P(E_1) = {10\over63}$, $P(E_2) = {10\over21}$, $P(E_1) = {20\over63}$
$P(B) = P(B|E_o)P(E_0)...P(B|E_3)P(E_3) = {1\over126}*1+{10\over63}*{3\over4} + {10\over21}*{1\over2} + {20\over63}*{1\over4} = 4/9$
And $P(A\cap B) = {10\over63}*{3\over4}*{1\over3} + {10\over21}*{1\over2}*{2\over3} + {20\over63}*{1\over4}*1 = 5/18$
And to end: $P(A|B) = P(A \cap B)/P(B) = 5/8$.