# Fundamentals of probability theory - conditional probability

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.

Question

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