Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer

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, http://en.wikipedia.org/wiki/Monty_Hall_problem

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.