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post-21174-12640358266089.jpg

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2/3

there are 3 possible blue marbles. 2 of which are in a box with one other blue marble.

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1/2 because if you pulled a blue marple then it must be one of the 2 last boxes where one has another blue marple and one has another yellow one

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Posted (edited) · Report post

1/3 because it is only possible to draw two blue marbles from one of the three boxes.

Wait...actually, magician's answer seems to make sense now...

Edited by phil
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It is not asking for the probability of having 2 blue balls at the same time It is rather asking after picking a blue ball what is the probability of having the other blue as well

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Ahh...nevermind. I get it now.

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Posted · Report post

.

100%

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Posted (edited) · Report post

Remember that P(A|B) (The probability of A, given that B has already occurred) = P(A and B)/P(B)

We just pulled a blue marble from a random box, which means the first marble we picked from a random box was blue. This is our given. We want the probability that the other one is blue GIVEN that at least one is blue. That is, we are looking for...

P(Second is blue GIVEN first was blue) = P(Second is blue AND first was blue)/P(First is blue) = P(Both are blue)/P(First is blue). This is equal to (1/3)/P(First is blue).

P(First is blue) = 1/3*P(First is blue from first box) + 1/3*P(First is blue from second box) + 1/3*P(First is blue from third box) = 1/3*0 + 1/3*1 + 1/3*1/2 = 1/2.

So our probability is (1/3)/(1/2) = 2/3.

Edited by Chuck
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Posted · Report post

50%

Keep it simple...

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The possibilities of picking up a blue ball are the first ball inn the second basket 2nd ball in the 2nd basket and in both case the other ball would be blue. Or you could pick up the second ball in the third basket in which case the other ball is not blue

so p( of other ball being blue) = 2/3

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Posted · Report post

Ignore the 1st box, you couldn't have pulled a blue ball out of that one.

This only leaves two outcomes:

A: You have already pulled a blue ball from the second box, leaving a blue ball to pull out.

B: You have already pulled a blue ball from the third box, leaving an orange ball to be pulled out.

They are the only two possible outcomes. The probabillity is 50%

Can't understand why people are getting in such a muddle with this one.

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Ignore the 1st box, you couldn't have pulled a blue ball out of that one.

This only leaves two outcomes:

A: You have already pulled a blue ball from the second box, leaving a blue ball to pull out.

B: You have already pulled a blue ball from the third box, leaving an orange ball to be pulled out.

They are the only two possible outcomes. The probabillity is 50%

Can't understand why people are getting in such a muddle with this one.

ah but then

there are two different balls in the second box u couldve picked the first or the second ball which gives you two different events

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Posted (edited) · Report post

No. Sorry. Both those events give the same result and are therefore the same.

ah but then

there are two different balls in the second box u couldve picked the first or the second ball which gives you two different events

Edited by gresleysteve
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Posted · Report post

50 50 or 50%

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.

If you have pulled a blue marble that eliminates the box with two orange marbles. That makes your chance of pulling a blue marble FROM THE SAME BOX 50%.

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Once you select a blue marble, you have already limited your next choice to only two boxes. Only one of these two has another blue marble so your probability of another blue one is 50%.

Edited by Tonybz
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Posted · Report post

Enumerate enumerate enumerate. That means count.

There are 6 ways to pull two marbles out.

1..2

2..1

3..4

4..3

5..6

6..5

we are given the first marble was blue

3..4

4..3

6..5

so we are down to 3 Possible Outcomes

how often is the second marble blue?

3..4

4..3

we have 2 Desired Outcomes

2/3

de-muddled?

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Posted · Report post

Forget the boxes. They're a distraction.

If we randomly select a box, and then randomly select a marble from that box, each marble has a 1/6 chance of being selected. So each marble is equally likely to be selected. If we select EITHER of the marbles that were in the box with two blues, then that marble's partner will also be blue. Since each marble is equally likely to be selected, there is a 2/3 chance that we selected one of those two marbles.

If you don't believe it, try an experiment. Roll a die a bunch of times, with each result representing the first marble chosen, and note the results:

1 or 2 = First box, orange (DISCARD the trial, since we were given that the first marble was blue)

3 or 4 = Second box, blue (First marble is blue, and its partner is guaranteed to be blue. Make a note here: HUGE SUCCESS.)

5 = Third box, orange (DISCARD, since first wasn't blue)

6 = Third box, blue (First marble is blue, but its partner is orange. FAILURE.)

Of the trials that weren't discarded, see how many are successful.

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The starting point for this problem is not the picture given with three boxes.

The problem states that the given condition is that one blue ball has been removed from a box. Thus the first box with two orange balls is eliminated and two boxes remain; one with a blue ball and one with an orange ball. Therefore, the probability that a blue ball will be drawn from one of the remaining boxes is 50%.

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There seems to be a lot of confusion over what is a relativly simple problem.

YOU HAVE ALREADY TAKEN A BLUE BALL OUT OF ONE BOX.

Therefore, it is not the box with two yellow balls in.

That leaves two boxes to think about.

One box had only one blue ball in, now it has none. the other had 2 blue balls in, now it has one. So, there's a 50-50 choice. The probability that THE OTHER BALL IN THE BOX YOU HAVE ALREADY SAMPLED is blue has to be 0.5; i.e. a half.

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This problem is very similar to the Monty Hall problem http://en.wikipedia.org/wiki/Monty_Hall_problem, which was also discussed in this forum here

This problem boils down to assigning the probability to each box of pulling the first blue marble out of that box. Assuming that the first marble is pulled absolutely at random and just happened to be blue, the probability of pulling a blue marble of the first box is 0. It's twice as likely to pull a blue marble from the second box than from the third, so the probabilities look like this:

Box #1: 0

Box #2: 2/3

Box #3: 1/3

You are picking a random marble, so each marble has equal probability to be picked and that probability is 1/6. 3 out of 6 marbles are blue, so there is 1/2 probability that the marble you pick will be blue. So, out of 6 possible choices, 3 choices lead to a blue marble. 2 out of 3 come from the box #2 and 1 comes from the box #3. Therefore, the probabilities are 2/3 and 1/3 respectively.

Now it's easy to see that after we pulled a blue marble the only way to have another blue marble is if we pulled it from the box #2 and the probability of that is 2/3.

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It is not asking for the probability of having 2 blue balls at the same time It is rather asking after picking a blue ball what is the probability of having the other blue as well

The probability of having 2 blue balls is significantly less than having one blue ball. TRUST ME!

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Let's say that there are more balls in some boxes. First box has 100 orange, second has 100 blue, and third has 1 blue and 99 orange. We randomly select a box, and then randomly select a ball from that box. It's blue. Is there a 50% chance that this ball came from box 3 and 50% from box 2?

For another approach, we'll use conditional probability again. Remember P(A|B) = P(A AND B)/P(B). That is, P(A given B) = P(A and B)/P(B). Our given is that the first ball chosen was blue. This probability, I think we can all agree, is 1/2 (3 blue out of 6). The probability that we want to figure out is the probability that the second ball is blue (which will happen IF AND ONLY IF we chose box 2), given that this first ball was blue. We had a 1/3 chance of choosing box 2. P(Second is blue GIVEN first was blue) = P(Second is blue AND first is blue)/P(first is blue) = P(Second box chosen)/P(first is blue) = (1/3)/(1/2) = 2/3.

Finally, someone already did this approach, but it's the easiest to follow, so I'll repost it. Just brute force it. There are six ways to choose the first marble, and then we can just examine the other marble after we pick the first one:

1-2 = discard (first wasn't blue)

2-1 = discard (first wasn't blue)

3-4 = success (both were blue)

4-3 = success (both were blue)

5-6 = failure (first was blue, second orange)

6-5 = discard (first wasn't blue)

Of the non-discarded trials, we have 2 successes out of 3, so our desired probability is 2/3.

I promise, the answer is 2/3.

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It seems that some people are getting this problem confused with the infamous game show problem. But there is a very key difference here. In tha game show problem you are told to pick from 3 doors, leading to the enumeration that many are proposing. But in this problem your first choice has already been given to you. You have picked a blue marble, which automatically eliminates the first box as a possibility. There could be another box with two orange marbles, or 5 other boxes, or 100 other boxes, but when the information is given that you have picked a blue marble, the problem limits you to the two boxes that have the blue marbles.

Or another way to think about it. If the third box also had 2 blue marbles, what would the answer be? It would be 100% because you know that a box with a blue marble has another blue marble in it, regardless of what's in the first box. Change one to orange and the probability becomes 1/2.

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It seems that some people are getting this problem confused with the infamous game show problem. But there is a very key difference here. In tha game show problem you are told to pick from 3 doors, leading to the enumeration that many are proposing. But in this problem your first choice has already been given to you. You have picked a blue marble, which automatically eliminates the first box as a possibility. There could be another box with two orange marbles, or 5 other boxes, or 100 other boxes, but when the information is given that you have picked a blue marble, the problem limits you to the two boxes that have the blue marbles.

Or another way to think about it. If the third box also had 2 blue marbles, what would the answer be? It would be 100% because you know that a box with a blue marble has another blue marble in it, regardless of what's in the first box. Change one to orange and the probability becomes 1/2.

Like Chuck said, the boxes are a distraction. You aren't selecting boxes, you are choosing marbles and you have a 2/3 chance of having selected a marble from the double blue box once you find the first marble is blue. Your alternate method of viewing the problem is flawed. Yes, you have a 100% chance if both boxes have 2 blue marbles, but it doesn't make sense for the probability to go to 50% if you change 1 out of 4 marbles. In your example, you started with 4 possible marbles that you selected first. After you changed 1 of the marbles to orange, that still leaves 3 marbles that you could have selected in your first pick. Just because two have the same value from the same box doesn't make them the same event. You're twice as likely to have picked a marble from box 2 in the first place.

Chances are this puzzle already exists in some form on this forum as I've seen it with three cards marked with Xs and there is a Wikipedia article on the puzzle with a slightly different casting as well, though the problem itself is identical. The article explains why Chuck et al. have the correct interpretation of the problem and also recognizes that the problem is similar to the Monty Hall problem and the boy and girl problem, both of which have been discussed at length on this forum. :rolleyes:

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