[Guest]

A bag contains 4 balls.

Two balls are drawn at random and are found to be white.

What is the probability that all balls are white?

[Guest]

I am assuming that the question should be interpreted in English and not some made up language that by coincidence uses English words but with different definitions that only the original poster understands.

Nice example. LOL.

[Guest]

Lol, thanks.

[Guest]

Ok if we make the assumption that each ball is either black or white with equal probability, then the probability that all 4 are white is the probability that the 2 in the bag are both white. There are 4 possible outcomes: WW, WB, BW, BB Only one case in 4 are they both white, so the answer is 0.25.

[superprismatic]

You therefore assume a uniform random variable from 0 to 4 and I assume a binomial random variable with parameters n=4, p=0.5;

Actually, I tried to say that, if you were to average over all distributions -- all equally

likely in the absence of information -- then you get the uniform distribution. So, in that

sense, I didn't just pick it out of the air. After all, that bag may have been filled by

someone who has a marked preference for pastel colors and added some white ones for balance.

Or, perhaps, it is a bag made by someone who was filling bags with balls of colors in

alphabetical order and he started this bag with Ultramarine (U) ,Violet (V), White (W)

but, since he couldn't come up with a color starting with X, he got frustrated and just

stuck another White (W) just to get four in there! I give each distribution (whether binomial

or even one conveying a very personal bias) equal weight.

[Guest]

0.375 I just multiplied stuff together, doing my probabilities different

[Guest]

I do have to agree with mmiguel1 here that not enough information is given in order to solve.

[Vipe195]

1/3

[Guest]

Actually, I tried to say that, if you were to average over all distributions -- all equally

likely in the absence of information -- then you get the uniform distribution. So, in that

sense, I didn't just pick it out of the air. After all, that bag may have been filled by

someone who has a marked preference for pastel colors and added some white ones for balance.

Or, perhaps, it is a bag made by someone who was filling bags with balls of colors in

alphabetical order and he started this bag with Ultramarine (U) ,Violet (V), White (W)

but, since he couldn't come up with a color starting with X, he got frustrated and just

stuck another White (W) just to get four in there! I give each distribution (whether binomial

or even one conveying a very personal bias) equal weight.

But how can you possibly average all possible distributions?

I suppose a symmetry argument?

If f(k) = P[k white balls initially]

then nothing distinguishes f(0) from f(1) from f(2) ... f(4) and hence they should all be equal?

I think it is hard to know what the average is.

You will end up with an infinite series.

Things will get complicated.

I think without a specification in the problem, we all have to draw something out of the air.

Edited March 22, 2010 by mmiguel1

[superprismatic]

I think it is hard to know what the average is.

You will end up with an infinite series.

Things will get complicated.

Yes, it is an infinite series. But since every permutation of every distribution is

included, each term in the average must be equal.

[Vipe195]

You guys are making this to complicated.

Edited March 22, 2010 by Vipe195

[Guest]

Yes, it is an infinite series. But since every permutation of every distribution is

included, each term in the average must be equal.

Alright, but you still give equal weights to all distributions which is an arbitrary decision.

Some distributions make more sense to assume than others, for example I think that the binomial and uniform distributions would make more sense than say the distribution corresponding to P[all black balls] = 1 and P[anything else] = 0 (which clearly cannot be true given that you draw two white balls).

But even then I think that combining the binomial and uniform distributions into an average would give an answer that is wrong rather than taking exclusively one or the other and getting a result that makes sense on a practical level.

So even if you weighted different distributions, I don't think it is best to combine them into an average.

Edited March 22, 2010 by mmiguel1

[superprismatic]

You guys are making this to complicated.

Spring has sprung

The grass is riz

I wonder where the birdies iz?

--- simple and to the point

[Guest]

I do agree that this thread could use some rest.

[superprismatic]

Alright, but you still give equal weights to all distributions which is an arbitrary decision.

Some distributions make more sense to assume than others, for example I think that the binomial and uniform distributions would make more sense than say the distribution corresponding to P[all black balls] = 1 and P[anything else] = 0 (which clearly cannot be true given that you draw two white balls).

But even then I think that combining the binomial and uniform distributions into an average would give an answer that is wrong rather than taking exclusively one or the other and getting a result that makes sense on a practical level.

So even if you weighted different distributions, I don't think it is best to combine them into an average.

You make a lot of good points. If you're not a mathematician, you should consider being one! Mathematics needs thinkers that challenge other people's ideas. You have a very admirable quality. I don't know if you are right in this case or not, but you approach the problem with gusto! I'll call a truce. I hope you accept my terms.

[Guest]

What on earth will Kewal think when he checks up on this thread .

[Vipe195]

There are three ways this could go-

1. 2 Non white Balls

2. 1 non white 1 white

3. 2 white

Therefore the probability of pulling two more whites is 1/3

[Guest]

You make a lot of good points. If you're not a mathematician, you should consider being one! Mathematics needs thinkers that challenge other people's ideas. You have a very admirable quality. I don't know if you are right in this case or not, but you approach the problem with gusto! I'll call a truce. I hope you accept my terms.

Of course! Nothing but friendly debate and you do bring many good arguments.

I am actually an electrical engineering student but I do love math.

Probability is fresh in my mind because I just took a final for a course devoted to probability.

Truce accepted.

[Guest]

i think ZERO- there is an infinite amount of colors there could be, couldn't there?

and... um... is it just me or did you make your title sound suuper-perverted on purpose? i mean, "white balls." it sounds like... ok im not gonna finish that sentece

[Vipe195]

Think of it this way.

X stand for all colors.

These are the possible outcomes-

1.2 x balls

2.1 x ball 1 white ball

3.2 white balls

So the answer would be 1/3

Edited March 23, 2010 by Vipe195

[Guest]

Think of it this way.

X stand for all colors.

This is how it would go-

1.2x balls

2.1x ball 1 white ball

3.2 white balls

So the answer would be 1/3

That requires that each of those cases is equally likely.

Unless you assume a contrived distribution for the white balls in the initial bag, those cases will not be equally likely.

What if an example distribution gives you the probabilities:

1. 0.25

2. 0.5

3. 0.25

Instead of

1. 0.33...

2. 0.33...

3. 0.33...

as you are assuming.

From only what you are considering, those probabilities can be nearly anything.

To make a better argument you must consider the probabilities for a given bag to have different numbers of white balls without the additional information of drawing two balls.

There many posts behind us that discuss this.

Edited March 23, 2010 by mmiguel1

[Vipe195]

Again you are making it too compilicated.

[Guest]

Again you are making it too compilicated.

I make it as complicated as it needs to be for it to be correct.

Not everything has a simple answer.

Assuming the answer is 1/3 because you can think of three possibilities left in the bag is not sound reasoning and if the probability actually were to be 1/3 it would not be due to your explanation.

I do agree with your approach in your response to teekayla though.

It doesn't matter what colors are possible. Only two categories of balls are of consequence: those that are white and those that are not.

Every mention of black balls before can be replaced with references to non-white balls.

Edited March 23, 2010 by mmiguel1

[Glycereine]

I make it as complicated as it needs to be for it to be correct.

Not everything has a simple answer.

Assuming the answer is 1/3 because you can think of three possibilities left in the bag is not sound reasoning and if the probability actually were to be 1/3 it would not be due to your explanation.

I do agree with your approach in your response to teekayla though.

It doesn't matter what colors are possible. Only two categories of balls are of consequence: those that are white and those that are not.

Every mention of black balls before can be replaced with references to non-white balls.

Even if there is equal probability for a ball to be white or some other color, it's not 1/3 because its not simply WN, WW, NN, it's WWWN, WWWW, WWNW, WWNN for the posibilities that still exist (out of the ones not already eliminated by the first 2 balls being white).

And the assumption that there is an equal probability for the balls to be white or not-white assumes only 2 colors of balls exist and that they exist in equal numbers... So yeah, not "too complicated" but as complicated as it needs to be as mmiguel mentions.

I honestly have to maintain it's flat out not solvable currently.

[Vipe195]

Since there is WWWN WWWW WWNW WWNN that would be 1/2

But the order doesn't count so WWWN and WWNW would count as one

that would make it WWWN/WWNW WWWW WWNN

Making me right with 1/3

Edited March 23, 2010 by Vipe195

[Guest]

Since there is WWWN WWWW WWNW WWNN that would be 1/2

But the order doesn't count so WWWN and WWNW would count as one

that would make it WWWN/WWNW WWWW WWNN

Making me right with 1/3

I'm sorry but you are wrong on two major levels.

The easiest to point out is like I said before, just because you count three things doesn't mean that each probability is 1/3.

Before you tell me I'm making things too complicated, think about this:

Flip two coins and count the number of heads.

There are three outcomes for the number of heads: 0 heads, 1 head, or 2 heads.

That does NOT mean that the probability of getting 0 heads is 1/3 even though there are three possibilities for the number of heads.

It is actually 1/4.

THE PROBABILITY OF ONE OF N OUTCOMES IS NOT NECESSARILY 1/N

There is a 50% probability to get tails on the first flip. For those times you do get tails, there is a 50% chance of getting tails again on the second flip.

0.5 * 0.5 = 1/4 and we multiply these events because we are interested in both of them occurring and they are independent (the result of the first flip does not affect the result of the second flip).

The other thing you are wrong about is that you are not taking into account enough information.

The probability of the remaining two balls being white depends on both the probabilities for different numbers of white balls being in the bag initially and on the fact that you drew two white balls already.

Your explanation does not take into account any of this information.

If you really want to understand what is going on, read the other posts in this thread.

Edited March 23, 2010 by mmiguel1