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I am unsure if anyone has seen this problem anywhere, but I haven't. This is an original question, and later I realized, with my inability to answer it, I do have a fundamental gap in my understanding of probability theory. Conditional probability to be precise.


1. Say I were to give you an urn, and I say that there are either 5 black balls
   and 4 white balls in it OR there are 4 black balls and 5 white balls in it,
   but I'm not sure how many.
What is the probability of drawing a white ball?

How different is the previous question from this one, which has a little more information.

2. There are 3 white balls and 2 black balls in the first box. There are 4 white
   and 4 black balls in the second box. A ball is randomly chosen from the
  first box and placed in the second box. A ball is then randomly chosen from
  the second box. What's the probability the chosen ball is white?

The solution to the question 2, is obviously P(B) = P(A) *(P(B/A) + P(B/A'))
= 23/45 with events A and B suitably defined (any adept mathematician,
should be able to define A and B)

Would you do the same to question 1, by assuming a probability of 1/2 to each case and calculate as

1/2*(5/9 + 4/9) = 1/2
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yes, i would do that. – Aang Mar 7 '13 at 18:37
I am not a mathematician (an engineer, in fact) and so could you explain to us non-mathematican readers what definition of $A$ and $B$ leads to $$P(B) = P(A)\left(P(B\mid A) + P(B\mid A^c)\right)?$$ – Dilip Sarwate Mar 7 '13 at 18:51
One could, although "equally likely" is not the only possible interpretation of "don't know." I don't know whether it will rain tomorrow, but the events "rain" and "no rain" are not equally likely. – André Nicolas Mar 7 '13 at 19:34
up vote 1 down vote accepted

Yes.(There is a typo in the 2nd Question's equation, btw.) Nevertheless, what does conditional probability say? You know the equation. But what it says essentially is the reason we add (or multiply) probabilities. (In the first class of your probability.)

You add probabilities of two events when both events can be done, separately. From here one can simplify operations by multiplying.(Try this- Now the conditional probability that you are getting is only because of adding all the seperate cases possible in which each case multiplying is obtained multiplying those 'two' terms(ex:P(A)P(B/A). You know why the multiplication is done here!.)

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