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March 5th, 2014, 08:11 PM   #1
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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?
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March 5th, 2014, 09:23 PM   #2
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Re: Marbles

Quote:
Originally Posted by shunya
In a bag, there is either a black or white marble.
With equal probability?
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March 6th, 2014, 12:11 AM   #3
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Re: Marbles

Quote:
Originally Posted by skipjack
Quote:
Originally Posted by shunya
In a bag, there is either a black or white marble.
With equal probability?
YES
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March 6th, 2014, 08:02 AM   #4
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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.
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March 6th, 2014, 10:02 AM   #5
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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):
  • The bag contains two white marbles[/*:m:c64e6ym4]
  • The bag contains one black and one white marble[/*:m:c64e6ym4]
Thus, when a white marble is drawn, there is a 50% chance that the marble in the bag is black.
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March 6th, 2014, 11:24 AM   #6
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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...
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March 6th, 2014, 12:53 PM   #7
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Quote:
Originally Posted by v8archie
Thus, when a white marble is drawn, there is a 50% chance that the marble in the bag is black.
No, because that would be the case if a white marble were chosen deliberately. The problem states that it happens to be white after being drawn at random. The possibilities you listed are equally likely.

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.
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March 6th, 2014, 04:31 PM   #8
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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.
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March 6th, 2014, 05:35 PM   #9
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Re: Marbles

Quote:
Originally Posted by Denis
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...
Wrong, dummy! 2/3 prob it's white...so 1/3 it's black

(1)(1/2) / (3/4) = 2/3
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