Three urns contain balls. Urn A contains 2 black balls and 2 red balls. Urn B contains 3 white balls and 1 black ball. Urn C contains 1 black, 1 white and 1 red ball.
An urn is chosen, and then two balls are removed one after the other.
The probability to chose urn A is 0.5, the probability to choose urn B is 0.25 and the probability to chose urn C is 0.25. in each urn any ball is equally likely to be removed.
Now I have to calculate the probability that at least one black ball is removed and the second ball removed is white.
In the former parts of this question, I already calculated the probability that at least one black ball is removed, and the probability is $ \frac{17}{24} $
and I calculated the probability that the second ball removes is white, the probability is $ \frac{13}{48} $.
Now I want to use the conditional probability formula and denote $ E_1 $ as the event that at least one balck ball removed, and $ E_2 $ as the event that the second ball removed is white.
Thus, $ \mathbb{P}\left(E_{1}\cap E_{2}\right)=\mathbb{P}\left(E_{1}|E_{2}\right)\cdot\mathbb{P}\left(E_{2}\right) $
I already calculated $ \mathbb{P}\left(E_{2}\right) $ so all I have to do is to calculate $ \mathbb{P}\left(E_{1}|E_{2}\right) $. given that $ E_2 $ occured, the only way at least one black ball is removed is if the first ball removed is black, also since we know that the second ball removed was white, it cannot be that urn A was picked. therefore the probability we want is the probability that urn B was picked multiplied by the probability that the first ball is black, plus the probability that urn C was picked multiplied by the probability that the first ball is black. That is :
$ \frac{1}{4}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3} = \frac{7}{47} $
now if we'll mulyiply it by the probability of $ E_2 $ we'll get $ \frac{7}{47}\cdot\frac{13}{48} $ which leads to the wrong answer.
I'll now present the right answer, but I'd like to know where the first way falls, Im new to this and I dont understand what I did wrong.
This is the right way:
without using the conditional probability formula, assuming the second ball is white, the only possibilities we are interested in are $\left(B,black,white\right),\left(C,black,white\right) $
That is the probability:
$ \frac{1}{4}\cdot\frac{1}{4}\cdot1+\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{16}+\frac{1}{24}=\frac{5}{48} $.
Thanks in advance.
Best Answer
Your first approach was wrong, because although you have correctly identified the probability of white being chosen (2nd) as 13/48, under this scenario, it is significantly more likely that the white ball came from urn B than urn C. Intuitively, this is because when urn B is selected, your chance of a white ball is 75% vs the 33.33% from urn C. Intuiting to the extreme, suppose that urn C had 100 balls, only 1 of which was white. Then if a white ball is picked, it is much more likely that it would have come from urn B.
This is the main problem with your first approach.
There is also an auxiliary problem.
Suppose that urn B is chosen, and it is assumed that the 2nd ball chosen is white. Then the chance that the 1st ball chosen is black is 1/3, not 1/4. This is because you have to eliminate one of the white balls in urn B from contention.
An easier way to intuit this is to pretend that the white ball is to be chosen 1st and the black ball is to be chosen 2nd. Then, the consideration re the last paragraph becomes more obvious.
Further, an easy way to see that this type of intuition is required is to pretend that urn B had two balls - 1 white and 1 black.
Then, under the assumption that Urn B was selected, and that the 1st ball chosen was white, the chances would be certainty rather than 1/2 that the 2nd ball chosen is black.
A similar error occurred with respect to your temporary consideration of Urn C being chosen.