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Friday Puzzle -- Another from Lewis Carroll

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Friday Puzzle -- Another from Lewis Carroll

#1

Friday Puzzle -- Another from Lewis Carroll

Alex Y

You are given an empty bag and a large bag of marbles that has an equal number or black and white marbles. Without looking, you reach into the bag and pick a marble and put it in the empty bag. You then put a white marble into that bag with the other marble.

Now you reach into the bag with two marbles and pull out a marble, which is white. What is the probability that the other marble in that bag is white?

Re: Friday Puzzle -- Another from Lewis Carroll

#2

Re: Friday Puzzle -- Another from Lewis Carroll

Larry Barrett

I think 2/3. It seems like Mr. Bayes could come to the rescue here but I can not construct the events properly.

I look at it this way:

There are 4 possible sequences, all equally likely.

1. First ball in the bag is W, second ball in the bag is W, draw the first ball and observe color (W), color of remaining ball = W.

2. First ball in the bag is W, second ball in the bag is W, draw the second ball and observe color (W), color of remaining ball = W.

3. First ball in the bag is B, second ball in the bag is W, draw the first ball and observe color (B), color of remaining ball = W.

4. First ball in the bag is B, second ball in the bag is W, draw the second ball and observe color (W), color of remaining ball = B.

Sequence 3 must be eliminated because the color of the drqwn ball must be W (by the statement of the problem).

Of the remaining 3 sequences that satisfy the conditions of the problem, 2 or the 3 have a W ball remaining in the bag. So probability the remaining ball is W =2/3.

Re: Friday Puzzle -- Another from Lewis Carroll

#3

Correct

Alex Y

Your solution is correct. I agree that a Bayesian analysis should work, but is more complex to set up correctly than what you did. I'll have to give that some thought.

Re: Friday Puzzle -- Another from Lewis Carroll

#4

Bayesian setup

Alex Y

It wasn't as hard to set up as I thought it would be.

A = The random Ball put in the bag was white

B = The ball drawn from the bag was white.

We want P(A|B)

Bayes tells us that P(A|B) = P(A)*P(B|A)/P(B)

P(A) = 1/2

P(B|A) = 1

P(B) = P(B|A)*P(A) + P(B|~A)*P(~A) = 1*.5+.5*.5 = 3/4

So, answer is 1/2 * 1 / (3/4) = 2/3

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