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Friday Puzzler -- bags of marbles

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Friday Puzzler -- bags of marbles

#1

Friday Puzzler -- bags of marbles

Alex Y

Bag A has 50 white and 50 black marbles.

Bag B has 100 white marbles.

Bag C starts out empty.

Without looking, you move one marble each from bag A and bag B into bag C.

You then remove one marble from Bag C and see that it is white.

What is the probability that the marble remaining in bag C is also white?

Re: Friday Puzzler -- bags of marbles

#2

Re: Friday Puzzler -- bags of marbles

Larry Barrett

Mr. Bayes, if I heard him correctly, said that the answer is 2/3.

Re: Friday Puzzler -- bags of marbles

#3

You heard Mr. Bayes correctly!


Re: Friday Puzzler -- bags of marbles

#4

Share, please

Alex Y

Would you mind sharing the Bayesian solution to this? I always have problems with the setup of a Bayesian equation, so end up going back to first principles like this:

I used upper case letters for the ball picked from the bag of all white balls and lower case for the ball picked from the bag with fifty-fifty mix. The equally likely possibilities are:

Picked first / Picked second

w / W

b / W

W / b

W / w

We know that we can eliminate the second possibility, leaving three possibilities, two of which include a white ball picked second.

Re: Friday Puzzler -- bags of marbles

#5

Re: Share, please

Larry Barrett

I also have trouble with setting up the Bayesian solution, but this is how I approached this problem.

Let A be the event of drawing two white balls from bag C, and let B be the event that the first ball drawn is a white ball.

So we want to compute P(A|B).

Bayes theorem says that P(A|B) = P(A)*P(B|A)/P(B)

Bag C could contain two white balls or it could contain one white and one black. Both are equally likely. It might help to think of bag C being either bag C1 with two white balls or bag C2 with one white ball, one black ball.

So P(A) is the probability of selecting two white balls, which would happen if you selected bag C1 with probability = 1/2.

P(B|A) is the probability of selecting one white ball, given that you already selected two white balls, so P(B|A) = 1.

P(B) is the probability of selected one white ball. This is the same as 1 - the probability of selecting one black ball. And the probability of selecting the black ball is is the probability of first selecting bag C2, which is 1/2, and then selecting the black ball from bag C2, which is also 1/2.

So P(B) = 1 - (1/2)*(1/2) = 3/4.

Therefore P(A|B) = ((1/2)*1)/3/4 = 4/6 = 2/3.

Re: Friday Puzzler -- bags of marbles

#6

Re: Share, please

Alex Y

Thanks. I fully understand, and maybe one day it will sink into my thick skull well enough for me to lay it out!

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