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**haikoumaths****Member**- Registered: 2009-07-16
- Posts: 3

"you know that you do not have Bag B (two black marble)"

but you also know you have one of two bags A and C.

If it is Bag A it contains one white ball and if it is Bag C it contains one black ball. There are not three balls to choose from, just the two. There is no "first" and "second" white ball because just as you have removed Bag B you have removed the so called "first" ball, it is the the ball you have in hand.

Why because had the balls in Bag A been numbered one and two and the balls in Bag C had one on the white and two on the black and then the ball removed from said bag been numbered one then you would know that which ever bag it came from (A or C) it must have left in the said bag a ball with number two on it.

Only by putting all three balls together in the said "same" bag can you have 2/3.

have I lost my marbles?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Welcome, haikoumaths!

I have just one thing to say,

Have lots of fun here!

Here's the question:

You have three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble.

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?

Maybe you'll understand the solution better if we list all six possible options.

1: You reach into Bag A and take out the first (white) marble. The other marble is white.

2: You reach into Bag A and take out the second (white) marble. The other marble is white.

3: You reach into Bag B and take out the first (white) marble. The other marble is black.

4: You reach into Bag B and take out the second (black) marble. The other marble is white.

5: You reach into Bag C and take out the first (black) marble. The other marble is black.

6: You reach into Bag C and take out the second (black) marble. The other marble is black

Of these, only the first three result in you taking out a white marble.

Out of those three options, two of them result in the other marble also being white.

If you're still not convinced, consider this rephrase:

You have three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble.

You pick a random bag and take out one marble.

What is the probability that the remaining marble from the same bag is the same colour?

Why did the vector cross the road?

It wanted to be normal.

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**haikoumaths****Member**- Registered: 2009-07-16
- Posts: 3

thank you

but the first action reduces the second action (that of determining the probable identity of the ball) to one bag (the same bag) which is either bag A or C. So that one bag which after the first action can only contain one ball which is either black or white.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The bag you selected is either A or C, but their probabilities aren't equal.

If you select A, then you're always going to pull out a white marble.

However, with C you'll pull out a black half the time and so only half of the C-selections are 'valid'.

Why did the vector cross the road?

It wanted to be normal.

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**haikoumaths****Member**- Registered: 2009-07-16
- Posts: 3

am I splitting hairs?

"You pick a random bag and take out one marble. It is a white marble."

That is the end of the first action.

Question "what is the probability that the remaining marble from the same bag is also white?"

we are asked to take it that the first action removes bag B therefor it is logical to take it that which ever bag the white ball did not come from is nolonger in he picture. so we have one bag, containing one ball which is either black or white. so the probability of that one ball (which we know is either black or white) being white is 1/2.

Hold on I think I might be wrong.

If you ask What is the probability of you choosing the bag containing the two white balls? the answer is 1/3

It seems to me to be like heads or tails, the fact that it landed five times heads in a row the next time (if it's a fair coin) the odds are still 1/2

*Last edited by haikoumaths (2009-07-18 00:53:17)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If you ask What is the probability of you choosing the bag containing the two white balls? the answer is 1/3

The question we really want to answer is

"What is the probability of choosing bag A, then drawing a white marble from it?"

The answer is still 1/3.

However, the same question for bag C has the answer of 1/6.

Therefore, pulling a white from A is twice as likely as pulling a white from C.

Why did the vector cross the road?

It wanted to be normal.

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**wyldr22****Member**- Registered: 2011-12-06
- Posts: 2

Not true. The question was "after pulling a white marble what is the probability?" Not, "what is the probability of choosing two white marbles?" The key is when is the question being asked. When it is said that a white marble has been pulled that reduces bag B out of the situation and a white marble out of the remaining bags. So it is down to a 50/50. Everyone has the correct math, just not in line with the wording. The wording eliminates the 1 out of 3 bag chance.

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**wyldr22****Member**- Registered: 2011-12-06
- Posts: 2

In reply #2 the original question and the rephrase are two different situations. The rephrase actually has the six possible outcomes that are listed above it. The original question by stating a white marble has been pulled, reduces the possible outcomes. Two bags are left with only one marble remaining in each. Since, it wasn't stated that we replace the marble. THis solution needs to be corrected on the site.

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