[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]

Or we could just take the question very literally. Since balls that are not white exist, then the probability of all balls being white is zero.

[Guest]

Well, consider the following argument. Let Pr(A|B) represent the

probability that of A given B is true). Further, let

A(0) be the statement that "the bag originally contained 0 white balls",

A(1) be the statement that "the bag originally contained 1 white ball",

A(2) be the statement that "the bag originally contained 2 white balls",

A(3) be the statement that "the bag originally contained 3 white balls",

A(4) be the statement that "the bag originally contained 4 white balls",

and B be the statement "2 white balls were randomly drawn from the bag".

We are asked to calculate Pr(A(4)|B). So,

(1) Pr(A(4)|B)=Pr(A(4)&B)/Pr(B)=Pr(B|A(4))*Pr(A(4))/Pr(B)

But,

(2) Pr(B)=Pr(B&A(0))+Pr(B&A(1))+Pr(B&A(2))+Pr(B&A(3))+Pr(B&A(4))

But, Pr(B&A(0))=0 and Pr(B&A(1))=0.

And so, (2) becomes

(3) Pr(B)=Pr(B|A(2))*Pr(A(2))+Pr(B|A(3))*Pr(A(3))+Pr(B|A(4))*Pr(A(4))

Using (3) to replace Pr(B) in (1) we get

(4) Pr(A(4)|B)=(Pr(B|A(4))*Pr(A(4)))/(Pr(B|A(2))*Pr(A(2))+Pr(B|A(3))*Pr(A(3))+Pr(B|A(4))*Pr(A(4)))

Now, we are told nothing about A(i) for i=2 to 4. So, we can't use that.

So, if we average all the ways we could assign velues to the A(i)s, we

get that they are all the same. We have no choice with no info about them.

Ignoring them (which is the same as setting them all to the same value),

we can divide them out of the numerator and denominator to get

(5) Pr(A(4)|B)=Pr(B|A(4))/(Pr(B|A(2))+Pr(B|A(3))+Pr(B|A(4)))

It is easy to see that Pr(B|A(2))=1/6, Pr(B|A(3))=1/2, and Pr(B|A(4))=1.

So,

(6) Pr(A(4)|B)=1/((1/6)+(1/2)+1)=6/(1+3+6)=6/10=0.6

Which is the solution to the OP.

0.6 is correct.

This question pertains to Baye's theorem.

Thanks a lot.

[Guest]

1/2?

[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?

I change my answer to

1/4

[

[Guest]

i'm thinking

that there's only four sets of the remaining balls so you must figure out the probability that you picked two white balls first based on each remainder set. Two other color balls left = 1/6 chance; one white, one other color= 1/2 (actually 6/12)chance twice; and both white = 1/1 chance

all of those time 1/4 equals an overall probability of...

13/24 Am I close?

[Guest]

as others have said i feel its unsolvable. there is no correct answer.

while its true you only have 3 cases;

1. non-white,non-white

2. non-white, white

3. white,white

as miguel pointed out there is no way to show that all three cases are equally likely.

[Guest]

as others have said i feel its unsolvable. there is no correct answer.

while its true you only have 3 cases;

1. non-white,non-white

2. non-white, white

3. white,white

as miguel pointed out there is no way to show that all three cases are equally likely.

It says next to the topic title that this was on an exam. I agree with others that it is unsolvable without making some assumptions, which makes it a poor exam question.

[Guest]

if they.ve already found to be white we know that they are white, so the probabilty is 1 right?

[Guest]

I hate "conditional probability"..... I would still like to claim that the fact that two white balls were drawn out doesn't change the initial probability that 4 white balls were put in the bag in the first place, assuming that the fact that you drew out two white balls was random and luck, you shouldn't be able to discard the other possibilities (BBBB, BBBW) and their permutations from your calculations. Except for that silly ocnditional probability convention....

I still hate it, its too messy.

By the way, there is plenty of info to solve this problem.

[Guest]

I hate "conditional probability"..... I would still like to claim that the fact that two white balls were drawn out doesn't change the initial probability that 4 white balls were put in the bag in the first place, assuming that the fact that you drew out two white balls was random and luck, you shouldn't be able to discard the other possibilities (BBBB, BBBW) and their permutations from your calculations. Except for that silly ocnditional probability convention....

I still hate it, its too messy.

By the way, there is plenty of info to solve this problem.

I had to reply to this. Whats wrong with conditional probability? It's extremely useful to use in real life and it is used in the statistical analysis of many things. To give a brief example of its usefulness:

Suppose there is a disease that 1 in ten thousand people have. There is a test you can take to see if you have it. The test is 99% accurate and only gives a false positive 1% of the time. You take the test and the results are positive. What is the probability that you have the disease?

about 1%...

[Guest]

I would still like to claim that the fact that two white balls were drawn out doesn't change the initial probability that 4 white balls were put in the bag in the first place....

By the way, there is plenty of info to solve this problem.

Your first statement is true, but if you have access to the information that two white balls were drawn, you don't care about the initial probability that 4 white balls were put into the bag as much as you care about the probability that 4 white balls were put in the bag GIVEN you drew 2.

P[4WB] is different from P[4WB | 2WD]

(B = bag, D = drawn)

You are right that P[4WB] doesn't change with drawing.

But the answer to the problem is P[4WB|2WD]

When questions state information like this, it is obvious that they are implying: what is the probability GIVEN all the additional information available to you (i.e. you draw 2 white balls).

Thus, the answer to the problem is conditional, and you cannot escape from using conditional probability if you really want to solve the problem.

Do not call something silly just because you fail to see it's usefulness.

There is a reason that conditional probability is a popular tool in science.

As for your second statement, I disagree. That's all I feel like writing about that.

Edited March 25, 2010 by mmiguel1

[Guest]

I'm struggling to see why people think there isn't enough info in this question.

If I said that you pick up a coin found in the street and toss it, what's the probability that you get heads? Would you say "That's an unanswerable question, as we do not know if it's a fair coin or not?" Or "That's an unanswerable question as we do not know if there is a general bias in the manufacture of all coins in the world that makes them favour one side or the other?" I hope that you would give the answer as 50%! Sure, there's a chance it's not a fair coin, but as we know no information about the coin, it is fair and reasonable to assume that it is either a fair coin, or that the proportion of coins biased towards heads is approximately the same as those biased towards tails, so the answer would still be 50%.

Similarly, if you pick up two die and roll them, what would you give as the probability of rolling 7? Would you say 1/6 or would you say that you can't give a probability yet, as you have yet to determine if both of the die are fair?

The same thing applies in this case. Given no constraints over what the initial selection of balls was from, it's right to assume that balls could be of any colour (of potentially millions of colours) and the selection of any one colour is equally as likely as the selection of any other. That is the generic question and thus supplies enough info to reach the answer. If the question wishes to impose additional constraints then it could do, but without those constraints we just widen up to the generic case.

When I first read the question I quite quickly got to the answer of 0.6 without using complicated probability, which is why I was surprised to see so much debate over it!

[Guest]

By the way, an interesting point to note is that most of the arguments saying there isn't enough information to go on are based on the principle that you need to be able to understand the probability of the bag containing a certain number of white balls before you consider the fact that you've already taken out two. If you had this information then giving the altered conditional probability would be fairly trivial.

The point with this question is that, although we can say that the probability that the bag has 4 white balls given that we've picked 2 is 0.6, we can not say anything about the probability that the bag contained 4 white balls before we picked any. (In fact, if we were to even attempt to answer that question, the best answer would be 0.) The point is that you don't need to know this (and don't even care about it) in order to answer the question.

[superprismatic]

When I first read the question I quite quickly got to the answer of 0.6 without using complicated probability, which is why I was surprised to see so much debate over it!

Please explain in detail how you came up with 0.6. I would like to see a simpler explanation than an application of Bayes Theorem.

[Guest]

Please explain in detail how you came up with 0.6. I would like to see a simpler explanation than an application of Bayes Theorem.

OK. Bear in mind that it's been about 12 years since I completed my maths degree, so whilst I remember a few things I will leave it to others to prove why they work.

In principle, how I approached this was as follows:

Given the information in the question, there are 3 possible scenarios:

1. There were initially 2 white and 2 non white balls.

2. There were initially 3 white and 1 non white balls.

3. There were initially 4 white balls.

If I use the term P(2W|SX) to mean the probability of picking 2 white balls given scenario X, then:

P(2W|S1) = 1/6

P(2W|S2) = 1/2

P(2W|S3) = 1

Now, probability of scenario 3 is P(2W|S3)/(P(2W|S1)+P(2W|S2)+P(2W|S3)) = 0.6

As I said above, I don't recall the exact reason why this works, but it is something that always stuck with me as it logically makes sense. I know the probability that I saw what I saw for each scenario, so I can now work out the probability that I am in any one of those scenarios. For completeness, the probability that there is one white ball left in the bag is 0.3 and the probability of none is 0.1.

You can quickly verify if that answer makes sense through logic as well. I.e. if there were only 2 white balls to start with then the chance of me picking both is quite slim, if there was a third white ball then the chance of picking 2 would go up, and if they were all white then I had to pick 2. Therefore you would expect the probability of scenario 3 to be greater than scenario 2, which in turn would be greater than scenario 1.

I think you probably gave the full proof of why this approach works in your complete answer. I remember principles and apply logic as it's quicker and easier!

[Guest]

The reason I think the initial probabilities are important can be explained by the following scenario:

Suppose I have 100 bags each of which contain exactly 4 balls. 1 of these bags contains 4 white balls. All the rest do not contain any white balls. I pick a bag at random. If the same question in the OP is asked now, the answer is obviously 1, not 0.6. So the original conditions do have an effect on the answer and need to be known.

I would like someone to reply to this as I am confused why people are saying otherwise.

[Guest]

I'm struggling to see why people think there isn't enough info in this question.

If I said that you pick up a coin found in the street and toss it, what's the probability that you get heads? Would you say "That's an unanswerable question, as we do not know if it's a fair coin or not?" Or "That's an unanswerable question as we do not know if there is a general bias in the manufacture of all coins in the world that makes them favour one side or the other?" I hope that you would give the answer as 50%! Sure, there's a chance it's not a fair coin, but as we know no information about the coin, it is fair and reasonable to assume that it is either a fair coin, or that the proportion of coins biased towards heads is approximately the same as those biased towards tails, so the answer would still be 50%.

Similarly, if you pick up two die and roll them, what would you give as the probability of rolling 7? Would you say 1/6 or would you say that you can't give a probability yet, as you have yet to determine if both of the die are fair?

The same thing applies in this case. Given no constraints over what the initial selection of balls was from, it's right to assume that balls could be of any colour (of potentially millions of colours) and the selection of any one colour is equally as likely as the selection of any other. That is the generic question and thus supplies enough info to reach the answer. If the question wishes to impose additional constraints then it could do, but without those constraints we just widen up to the generic case.

When I first read the question I quite quickly got to the answer of 0.6 without using complicated probability, which is why I was surprised to see so much debate over it!

Ok, with a coin, there is really only one distribution that is the most clear and obvious to choose from, being a Bernoulli distribution with probability 50% for success. With a die there is also only one probability distribution that is most clear and obvious to choose from, the number being a uniform discrete random variable from 1 to 6.

With putting colored balls into a bag the most clear and obvious distribution is not as apparent.

I would say it is reasonable to interpret the question as pertaining to only black and white balls because of how often questions like that appear, as well as more colors.

You suggest that there are potentially millions of colors and that the selection of any one color is as likely as the selection of any other color. This assumption does not lead to a probability of 0.6

If there are N colors and each color is equally likely as any other, then the probability of any particular color is 1/N.

Using your notation, you write

P(2W|S3)/(P(2W|S1)+P(2W|S2)+P(2W|S3))

This is actually incomplete.

Going back to the definitions,

P(S3|2W) = P(2W|S3)*P(S3)/P(2W) = P(2W|S3)*P(S3)/(P(2W|S1)*P(S1) + P(2W|S2)*P(S2) + P(2W|S3)*P(S3))

This actually isn't strictly correct either because S1, S2, S3 aren't collectively exhaustive. However in this case it does not matter because P(2W| the remaining events) = 0

To get your answer, neida, you assume that

P(S3) = P(S1) = P(S2) and divide each of these out.

I will show in a second that this step is not consistent with your assumption.

Your assumption is that any of N colors is equally likely.

Thus the probability of a given ball being white is 1/N and the probability of a given ball not being white is (N-1)/N.

Thus for scenario 1, the probability of getting 2 white and 2 non-white is

(4 choose 2) *(1/N)^2 *((N-1)/N))^2 = 6*(N-1)^2/N^4

For scenario 2,

the probability of getting 3 white and 1 non-white is

(4 choose 3)*(1/N)^3*(N-1)/N = 4*(N-1)/N^4

For scenario 3,

the probability of getting 4 white is

(1/N)^4

Plugging this into the formula and multiplying by N^4 in the numerator and denominator gives

P(2W|S3)/(P(2W|S1)*6*(N-1)^2 + P(2W|S2)*4*(N-1) + P(2W|S3))

Taking the limit as N goes to infinity as you seem to have been hinting at gives the probability to be zero.

This seems like too trivial of an answer to me.

Although Kewal disagrees with me, I think the best assumption when nothing is stated is to assume black and white balls with equal probability.

Just for interest's sake, I will continue what I was doing above, but will assume now that N is finite, with N=2 giving the answer I would argue.

Neida is right about these, which have also been shown numerous times in previous posts.

P(2W|S1) = 1/6

P(2W|S2) = 1/2

P(2W|S3) = 1

Substitution gives

P(S3|2W) = 1/((N-1)^2+2*(N-1)+1)

Substituting N=2 gives the answer 1/4.

To get 0.6, you must instead assume that P(S1) = P(S2) = P(S3)

That is you are equally likely to get 4 white balls as you are to get 3 white balls as you are to get 2 white balls.

You may extrapolate also to say it is also equiprobable to get 1 or 0 white balls, but that does not really matter.

Let us assume that for now it is equiprobable to get 0, 1, 2, 3, or 4 white balls.

Such a distribution gives an answer of 0.6.

There are 5 of these cases, and because they are mutually exclusive and collective exhaustive, the sum of the 5 equal probabilities gives 1. Each probability must therefore be 1/5 or 20%.

Imagine this distribution:

20% 0 whites

20% 1 white

20% 2 whites

20% 3 whites

20% 4 whites

How would you generate a bag with such a probability?

You cannot simply uniformly randomly pick out of N colors and put whatever you get in the bag.

You most likely would need to do some other random process which gives 5 different results with equal probability and from the result of that process, specifically count how many white balls correspond to the result and put them in the bag for the problem.

That does not seem like a natural thing for the solver to assume would happen in this problem.

I think that the solver definitely should assume that there are N equiprobable colors, in which case the answer is

1/((N-1)^2+2*(N-1)+1)

which is also

1/N^2

Edited March 26, 2010 by mmiguel1

[Guest]

The reason I think the initial probabilities are important can be explained by the following scenario:

Suppose I have 100 bags each of which contain exactly 4 balls. 1 of these bags contains 4 white balls. All the rest do not contain any white balls. I pick a bag at random. If the same question in the OP is asked now, the answer is obviously 1, not 0.6. So the original conditions do have an effect on the answer and need to be known.

I would like someone to reply to this as I am confused why people are saying otherwise.

Agreed

[Guest]

OK. Bear in mind that it's been about 12 years since I completed my maths degree, so whilst I remember a few things I will leave it to others to prove why they work.

In principle, how I approached this was as follows:

Given the information in the question, there are 3 possible scenarios:

1. There were initially 2 white and 2 non white balls.

2. There were initially 3 white and 1 non white balls.

3. There were initially 4 white balls.

If I use the term P(2W|SX) to mean the probability of picking 2 white balls given scenario X, then:

P(2W|S1) = 1/6

P(2W|S2) = 1/2

P(2W|S3) = 1

Now, probability of scenario 3 is P(2W|S3)/(P(2W|S1)+P(2W|S2)+P(2W|S3)) = 0.6

As I said above, I don't recall the exact reason why this works, but it is something that always stuck with me as it logically makes sense. I know the probability that I saw what I saw for each scenario, so I can now work out the probability that I am in any one of those scenarios. For completeness, the probability that there is one white ball left in the bag is 0.3 and the probability of none is 0.1.

You can quickly verify if that answer makes sense through logic as well. I.e. if there were only 2 white balls to start with then the chance of me picking both is quite slim, if there was a third white ball then the chance of picking 2 would go up, and if they were all white then I had to pick 2. Therefore you would expect the probability of scenario 3 to be greater than scenario 2, which in turn would be greater than scenario 1.

I think you probably gave the full proof of why this approach works in your complete answer. I remember principles and apply logic as it's quicker and easier!

When you memorize rules, you should also memorize when they are applicable.

Your rule is only applicable as an equation when P[s1]=P[s2]=P[s3]

You also shouldn't discredit other people who took the extra time to derive the answer from a more general rule than you bothered to use.

That is the better demonstration of remembering principles and applying logic.

Edited March 26, 2010 by mmiguel1

[Guest]

OK. Bear in mind that it's been about 12 years since I completed my maths degree, so whilst I remember a few things I will leave it to others to prove why they work.

In principle, how I approached this was as follows:

Given the information in the question, there are 3 possible scenarios:

1. There were initially 2 white and 2 non white balls.

2. There were initially 3 white and 1 non white balls.

3. There were initially 4 white balls.

If I use the term P(2W|SX) to mean the probability of picking 2 white balls given scenario X, then:

P(2W|S1) = 1/6

P(2W|S2) = 1/2

P(2W|S3) = 1

Now, probability of scenario 3 is P(2W|S3)/(P(2W|S1)+P(2W|S2)+P(2W|S3)) = 0.6

As I said above, I don't recall the exact reason why this works, but it is something that always stuck with me as it logically makes sense. I know the probability that I saw what I saw for each scenario, so I can now work out the probability that I am in any one of those scenarios. For completeness, the probability that there is one white ball left in the bag is 0.3 and the probability of none is 0.1.

You can quickly verify if that answer makes sense through logic as well. I.e. if there were only 2 white balls to start with then the chance of me picking both is quite slim, if there was a third white ball then the chance of picking 2 would go up, and if they were all white then I had to pick 2. Therefore you would expect the probability of scenario 3 to be greater than scenario 2, which in turn would be greater than scenario 1.

I think you probably gave the full proof of why this approach works in your complete answer. I remember principles and apply logic as it's quicker and easier!

Thanks a lot for explaining in a simple way.

[Guest]

Thanks a lot for explaining in a simple way.

A simple way that is not even consistent with his/her initial assumption that any color is equally likely and that there are millions of colors.

As I showed earlier that assumption gives an answer of 1/N^2 for N different colors.

So neida doesn't even know what assumptions he is making to get an answer of 0.6.

The only justification you guys provide is that it is simple and feels logical, and you ignore even dealing with the distribution of the white balls.

What you forgot in your equation is that there are weights on every term you use.

If you say P[X|S3], you instead needed to say P[X|S3]*P[s3].

The only way to ignore the weighting factors is to make all weights equal, and that is what you are assuming, whether it makes the physical sense or not, and whether you like it or not.

At least superprismatic gave a good reason for getting this answer, and acknowledged he made the assumption equivalent to making the number of white balls a uniform random variable.

This is the most frustrating thread I've come across.

[Guest]

Let's see if this answer is simpler (and as correct)... This gives the same answer as the more complicated analysis that precedes it.

A bag that has at least 2 white balls may come in 3 distinguishable proportion of balls.

It may come in...

1) W W W W

2) W W W O

3) W W O O

Where W means white balls and O are non-whites. And no other since there has to be at least 2 white balls, and order doesn't matter.

Let's write numbers on the 4 balls, so that each ball carries a different number.

For case 1) there are 6 choose 2 = 4 ways of picking 2 white balls carrying different numbers.

For case 2) there are 3 choose 2 = 3 ways of picking 2 white balls.carrying different numbers.

For case 3) there are 2 choose 2 = 1 ways of picking 2 white balls (carrying different numbers)

For bags of 4 balls, known to having at least 2 white balls, there are 10 different ways of picking 2 white balls, indeed.

The probability for case 1), all white balls is then 6/10 = 0.6 .

Edited March 27, 2010 by marsupialsoup

[Guest]

Let's see if this answer is simpler (and as correct)... This gives the same answer as the more complicated analysis that precedes it.

A bag that has at least 2 white balls may come in 3 distinguishable proportion of balls.

It may come in...

1) W W W W

2) W W W O

3) W W O O

Where W means white balls and O are non-whites. And no other since there has to be at least 2 white balls, and order doesn't matter.

Let's write numbers on the 4 balls, so that each ball carries a different number.

For case 1) there are 6 choose 2 = 4 ways of picking 2 white balls carrying different numbers.

For case 2) there are 3 choose 2 = 3 ways of picking 2 white balls.carrying different numbers.

For case 3) there are 2 choose 2 = 1 ways of picking 2 white balls (carrying different numbers)

For bags of 4 balls, known to having at least 2 white balls, there are 10 different ways of picking 2 white balls, indeed.

The probability for case 1), all white balls is then 6/10 = 0.6 .

I made a major error.

For case 1) there are 4 choose 2 = 6 ways of picking 2 white balls carrying different numbers.

For case 2) there are 3 choose 2 = 3 ways of picking 2 white balls carrying different numbers.

For case 3) there are 2 choose 2 = 1 ways of picking 2 white balls (carrying different numbers)

[superprismatic]

I made a major error.

For case 1) there are 4 choose 2 = 6 ways of picking 2 white balls carrying different numbers.

For case 2) there are 3 choose 2 = 3 ways of picking 2 white balls carrying different numbers.

For case 3) there are 2 choose 2 = 1 ways of picking 2 white balls (carrying different numbers)

I think we all noticed that that was a typo! For your information, mmiguel1 asserts that the three cases have different probabilities, thus causing the answer to be 0.25 rather than 0.60. I say this so you can appreciate the disagreement.

[Guest]

I think we all noticed that that was a typo! For your information, mmiguel1 asserts that the three cases have different probabilities, thus causing the answer to be 0.25 rather than 0.60. I say this so you can appreciate the disagreement.

I can accept the answer being 0.6 as long as there is a good reason, like yours.

I don't appreciate explanations that are both inconsistent and condescending though.

Really this argument shouldn't be a hot point of debate after this long.

I realize that much of this dragging along is my fault for so eagerly debating.

I say everyone should walk away with whatever conclusion they like best.