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A bag contains one billiard ball. It can be white or black (with equal probability). We put a white ball inside the bag (so now there are 2 balls in the bag).

Now we take one ball from the bag. It turns out the the ball we took is white.

Question: What is the probability that the ball left in the bag is black?

My opinion: There is $1/2$ probability that there is white ball in a bag, then we add white ball to the bag so the probability increases to $3/4$. Now if we take a ball from it and it turns to be white it means... I do not know what it means.

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up vote 1 down vote accepted

This is a conditional probability question. Let A be the event that the ball left is black, and let B be the event that the ball we draw is white. We want to find $P(A|B)$ or the probability that the remaining ball is black given the ball we drew is white.

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1/4}{3/4}=1/3$

Note: $P(B)=3/4$ as you showed in the question, and $P(A\cap B)=\frac{1}{2}(\frac{1}{2})$ since we have a $1/2$ chance of originally puting in a black ball, and then a $1/2$ chance of drawing the white ball.

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Hint: use Bayes formula. You have $P(W_1) = P(B_1) = 1/2$ and $P(W | W_1) = 1$, $P(W | B_1) = 1/2$. Now, you need to find $P(W_1 | W)$. I think you can do it.

It's a variant of Monty Hall problem,

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Excuse me. What is $P(W)$ and what is $W_1$, $W$,$B_1$? – Yoda May 20 '14 at 20:54
Is Peter Woolfitt right(below)? – Yoda May 20 '14 at 20:56

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