March 5th, 2014, 08:11 PM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  Marbles
In a bag, there is either a black or white marble. Without looking, you insert a white marble into this bag. Then you randomly pick a marble from this bag and it turns out to be white. What is the probability that the marble left in the bag is black? 
March 5th, 2014, 09:23 PM  #2  
Global Moderator Joined: Dec 2006 Posts: 21,036 Thanks: 2274  Re: Marbles Quote:
 
March 6th, 2014, 12:11 AM  #3  
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  Re: Marbles Quote:
 
March 6th, 2014, 08:02 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,036 Thanks: 2274 
There are three equally likely possibilities: (1) the only white marble is taken out, leaving a black marble, (2) the added white marble is taken out, leaving the original white marble, and (3) the original white marble is taken out, leaving the added white marble. Hence the required probability that the remaining marble is black is 1/3. 
March 6th, 2014, 10:02 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra  Re: Marbles
The way I read the question, there are only two marbles. One in the bag (which is either black or white) and one not in the bag (white). So, when drawing there are two possibilities (equally likely):

March 6th, 2014, 11:24 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Marbles
Well, I say 1/4 only... Half the time, there will be a black in the bag And half that time a white will be pulled out... 
March 6th, 2014, 12:53 PM  #7  
Global Moderator Joined: Dec 2006 Posts: 21,036 Thanks: 2274  Quote:
For the first case, when a marble is chosen at random, the probability it is white is only 1/2. For the second case, the marble then chosen at random has probability 1 of being white. The required probability is therefore (1/2)/(1/2 + 1) = 1/3. The method that Denis gave shows that the probability the marble chosen for removal was black was 1/4. Hence the probability that is was white was 3/4, comprising 1/2 corresponding to the case where both marbles in the bag are white and 1/4 corresponding to the case where the remaining marble is black. The required probability is therefore (1/4)/(3/4) = 1/3, as before.  
March 6th, 2014, 04:31 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra  Re: Marbles
Yeah, after I wrote it I was thinking "Monty Hall", but since one can't edit posts on here I just left it.

March 6th, 2014, 05:35 PM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Marbles Quote:
(1)(1/2) / (3/4) = 2/3  

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