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1) If we know we have a white marble chosen already then we had either bag A or C.

One of the bags has left a while marble and the other a black marble. So the probability

of choosing a bag and getting a white marble is 1/2. The probability that the remaining

marble in the bag is white is the same as choosing one of the remaining bags and then

picking the marble out of it. P(White)=P(choose bag A)*P(marble in A is white) = (1/2)*1=1/2.

P(Black)=P(choose bag C)*P(marble in C is black) = (1/2)*1 = 1/2.

2) If we know that we have bag A or bag C, but haven't gotten a marble out yet, then the

probability of getting a white marble by choosing bag A and then choosing a marble out

of it is P(W) = (1/2)*1 = 1/2. With bag C it is P(W) = (1/2)*(1/2) = 1/4 and

P(B) = (1/2)*(1/2) = 1/4. So the total probability for P(W)= 1/2 + 1/4 = 3/4 and P(B)=1/4.

In 2) the probability of getting a white marble is three times that of getting a black one. The ratio

of 3/4 to 1/4 is 1/3. Perhaps this is where the 1/3 comes from. But this 1/3 is not a probability,

it is a ratio of two probabilities.

To settle the dilemma perhaps we should average the 1/2 and 3/4 to get 5/8.

Or average 1/2 and 2/3 to get 7/12. Or average 2/3 and 3/4 to get 17/24.

I was once told by someone deep into probabilities to "Always look closely at the SAMPLE

SPACES." Typically if you change the sample space, then you change the probabilities.

The sample space in 1) is {{W},{B}} knowing we have chosen a W out of bag A or C.

The sample space in 2) is {{W,W},{W,B}} knowing we have one of these two bags to pick from.

In 1) we know bag A or bag C was chosen since we also chose a white marble from one of them.

In 2) we just have that bag A or Bag C are our options but nothing has been chosen from them.

But I'm not sure this is the same question posed in an earlier thread for which there was a

name attached ( so-and-so paradox?). Is this the same question?

Answer is really 1/2, not 2/3. Note that you are picking a "random bag" not marbles. So one bag has remaining white marble, one doesn't. The other bag is not a possibility with 2 black marbles. So, 1/2.

No, you're wrong. You pick a random bag, but you already know the result of the pick from your random bag. You've passed the random bag stage (thus why we can eliminate bag b as a possibility) and are already at picking a white marble. There are three white marbles, each with the same probability of being picked. 2/3 white marbles are with another white marble, therefore 2/3 of the time you'll draw a second white marble.

As usual these probabilities are easier to look at once you make the example absurd.

Now there are 1000 bags. 998 of them have two black marbles, 1 bag has two white marbles, 1 bag has one of each.

Even though the problem states we picked a random bag, if it also tells us we got a white marble out of it we know that it was one of two bags. For the sake of the problem the other 998 bags might as well never existed. We know that out of the 2000 marbles, we are only looking at 3. 2 of those are with each other, the last one is with a black marble. p=2/3

]]>You pick a random bag and take out one marble.

It is a white marble.

What is the probability that the remaining marble from the same bag is also white?

The Solution . . .

2/3 (not 1/2)

You know that you do not have Bag B (two black marbles) so there are three possibilities

You chose Bag A, first white marble. The other marble will be white

You chose Bag A, second white marble. The other marble will be white

You chose Bag C, the white marble. The other marble will be black

So 2 out of 3 possibilities are white.

Why not 1/2? You are selecting marbles, not bags."""

Answer is really 1/2, not 2/3. Note that you are picking a "random bag" not marbles. So one bag has remaining white marble, one doesn't. The other bag is not a possibility with 2 black marbles. So, 1/2.

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