[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]

zero

[Guest]

1/x where x = a lengthy argument about the infinite number of colors, and the deviousness of the poster and/or the person who added the balls to the bag

[Guest]

.25

[Guest]

1/4 or 0.25

[Guest]

1/4

Assuming that the balls are initially placed in the bag completely at random,

there is 1/16 chance of having 4 white balls

1/4 chance of having 3 white balls (4 choose 3 times 1/16)

3/8 chance of having 2 white balls (4 choose 2 times 1/16)

If there are 2 white balls the probability of drawing 2 is 2/4 * 1/3 = 1/6

If 3 white balls, drawing prob is 3/4 * 2/3 = 1/2

If 4, drawing prob is 1

Ans = 1*1/16/(1*1/16 + 1/2*1/4 + 1/6*3/8) = 1/(1 + 2 + 1) = 1/4

[Guest]

1/4

Assuming that the balls are initially placed in the bag completely at random,

there is 1/16 chance of having 4 white balls

1/4 chance of having 3 white balls (4 choose 3 times 1/16)

3/8 chance of having 2 white balls (4 choose 2 times 1/16)

If there are 2 white balls the probability of drawing 2 is 2/4 * 1/3 = 1/6

If 3 white balls, drawing prob is 3/4 * 2/3 = 1/2

If 4, drawing prob is 1

Ans = 1*1/16/(1*1/16 + 1/2*1/4 + 1/6*3/8) = 1/(1 + 2 + 1) = 1/4

Not correct so far.

[superprismatic]

.6

[Guest]

I agree with VAHID, with this initial conditions the bag can contain your testicles.

[Guest]

4 balls in bag

2 chosen at random

they are white

so they must all be white

probabilty 1

Edited March 22, 2010 by acrab

[Guest]

there are two possible outcomes for each ball.

we already know that the first two are white and the remaining two can either be white (w) or non-white (n). therefore, there are 3 remaining possible outcomes (order doesn't matter) => {wn,nn,ww}. only the ww outcome satisfies "all balls being white" therefore the probability is 1/3.

[Guest]

The problem does not provide a way to define the probability model.

Because it does not, assumptions must be made by the solver.

I assumed that when the balls were initially placed in the bag, a black ball was equally as likely as a white ball.

This makes the number of white balls a binomial random variable with parameters n=4 and p=1/2, i.e.

If X = number of white balls originally placed in the bag, then

p[X=k] = 4!/(k!*(4-k)!)*(1/2)^k*(1/2)^(4-k) = 4!/(k!*(4-k)!)*(1/16)

p[X=0] = 1/16

p[X=1] = 1/4

p[X=2] = 3/8

p[X=3] = 1/4

p[X=4] = 1/16

Note that it must be true that p[X=0] + p[X=1] + p[X=2] + p[X=3] + p[X=4] = 1

In this case:

W = white balls

D = drawn

B = in bag

Bayes' Rule

P[4WB | 2WD] = P[2WD | 4WB] * P[4WB] / P[2WD]

P[2WD | 4WB] = 1

P[4WB | 2WD] = P[4WB]/P[2WD]

By the total probability theorem:

P[4WB | 2WD] = P[4WB]/(P[2WD|0WB]*P[0WB]+P[2WD|1WB]*P[1WB] + P[2WD|2WB]*P[2WB] + P[2WD|3WB]*P[3WB] + P[2WD|4WB]*P[4WB])

It is easy to see that:

P[2WD|0WB] = 0

P[2WD|1WB] = 0

P[2WD|2WB] = 2/4*1/3 = 1/6

P[2WD|3WB] = 3/4*2/3 = 1/2

P[2WD|4WB] = 4/4 * 3/3 = 1

Then P[4WB|2WD] = P[X=4]/(0*P[X=0] + 0*P[X=1] + 1/6*P[X=2] + 1/2*P[X=3] + 1*P[X=4])

= 1/16/(1/16 + 1/8 + 1/16) = 1/4

You say this answer is incorrect.

If it is incorrect then it must be incorrect due to my assumption that when the bag was originally generated a black ball was equally as likely as a white ball. It cannot be subsequent calculations because the total probability theorem and bayes' rule are always valid.

The assumption I made seems to me to be the most natural assumption to make.

You must provide us a way to define the probability mass function of the number of white balls originally in the bag.

i.e. provide f(k) or a way to calculate it where P[X=k] = f(k) and X is the number of white balls initially in the bag.

For example, you might tell us to assume that f(k) = 1/5 i.e. having any number of white balls is equally likely as any other number of white balls.

In that case P[4WB|2WD] = 1/5/(1/30 + 1/10 + 1/5)= 1/(1/6 + 1/2 + 1) = 6/(1+3+6) = 0.6 as I believe superprismatic reasoned.

To generate a bag full of white balls with such a probability mass function is different than to arbitrarily choose to put in the bag a black or white ball with equal probability 4 different times (binomial random variable mentioned above).

You could spin a arrow/spinning wheel with 5 areas corresponding to the numbers 0,1,2,3,4 in a uniformly random way and then put in that many white balls into the bag or something like that.

Edited March 22, 2010 by mmiguel1

[Glycereine]

there are two possible outcomes for each ball.

we already know that the first two are white and the remaining two can either be white (w) or non-white (n). therefore, there are 3 remaining possible outcomes (order doesn't matter) => {wn,nn,ww}. only the ww outcome satisfies "all balls being white" therefore the probability is 1/3.

Concur.

I first went down the road of figuring out the probability of 2 white balls being drawn with each of the following sets of balls WWNN, WWWN, WWWW in the bag. Then I realized that has nothing to do with the probability of the other balls being white that are left. Unless some extra information is given we have to assume there is just as much chance for the remaining balls to be white as non-white. So Dawndada's answer is correct with what information we have.

[Guest]

with that information we have got, there is infinite number of solution, so every answer is correct [0;1]

i think the most obvious is to make easy your life

and assume that the bag can contain balls with any color with equal probability, so the solution is 0;

this is a bad riddle

Edited March 22, 2010 by det

[Guest]

The problem does not provide a way to define the probability model.

Because it does not, assumptions must be made by the solver.

I assumed that when the balls were initially placed in the bag, a black ball was equally as likely as a white ball.

This makes the number of white balls a binomial random variable with parameters n=4 and p=1/2, i.e.

If X = number of white balls originally placed in the bag, then

p[X=k] = 4!/(k!*(4-k)!)*(1/2)^k*(1/2)^(4-k) = 4!/(k!*(4-k)!)*(1/16)

p[X=0] = 1/16

p[X=1] = 1/4

p[X=2] = 3/8

p[X=3] = 1/4

p[X=4] = 1/16

Note that it must be true that p[X=0] + p[X=1] + p[X=2] + p[X=3] + p[X=4] = 1

In this case:

W = white balls

D = drawn

B = in bag

Bayes' Rule

P[4WB | 2WD] = P[2WD | 4WB] * P[4WB] / P[2WD]

P[2WD | 4WB] = 1

P[4WB | 2WD] = P[4WB]/P[2WD]

By the total probability theorem:

P[4WB | 2WD] = P[4WB]/(P[2WD|0WB]*P[0WB]+P[2WD|1WB]*P[1WB] + P[2WD|2WB]*P[2WB] + P[2WD|3WB]*P[3WB] + P[2WD|4WB]*P[4WB])

It is easy to see that:

P[2WD|0WB] = 0

P[2WD|1WB] = 0

P[2WD|2WB] = 2/4*1/3 = 1/6

P[2WD|3WB] = 3/4*2/3 = 1/2

P[2WD|4WB] = 4/4 * 3/3 = 1

Then P[4WB|2WD] = P[X=4]/(0*P[X=0] + 0*P[X=1] + 1/6*P[X=2] + 1/2*P[X=3] + 1*P[X=4])

= 1/16/(1/16 + 1/8 + 1/16) = 1/4

You say this answer is incorrect.

If it is incorrect then it must be incorrect due to my assumption that when the bag was originally generated a black ball was equally as likely as a white ball. It cannot be subsequent calculations because the total probability theorem and bayes' rule are always valid.

The assumption I made seems to me to be the most natural assumption to make.

You must provide us a way to define the probability mass function of the number of white balls originally in the bag.

i.e. provide f(k) or a way to calculate it where P[X=k] = f(k) and X is the number of white balls initially in the bag.

For example, you might tell us to assume that f(k) = 1/5 i.e. having any number of white balls is equally likely as any other number of white balls.

In that case P[4WB|2WD] = 1/5/(1/30 + 1/10 + 1/5)= 1/(1/6 + 1/2 + 1) = 6/(1+3+6) = 0.6 as I believe superprismatic reasoned.

To generate a bag full of white balls with such a probability mass function is different than to arbitrarily choose to put in the bag a black or white ball with equal probability 4 different times (binomial random variable mentioned above).

You could spin a arrow/spinning wheel with 5 areas corresponding to the numbers 0,1,2,3,4 in a uniformly random way and then put in that many white balls into the bag or something like that.

I have to agree with mmiguel1. With the information provided, we must assume that the chance of a white ball and the chance of a non-white ball being placed in the bag initially is the same. If you said the 4 balls could be any color and that every color has an equal chance, that would push the probability very close to 0%.

[Guest]

... 66.67%, here's the logic

In absence of any information about the original ball selection, I would think that the only know fact "that two random balls were white" should be used as a hint of how the balls were initially selected. Therefore, I only can assume three scenarios and will give them same weight (but not probability):

A) all four balls are white

B) three balls are white

C) two balls are white.

Now I want to know for each scenario, what is the probability of the event (two random balls were white):

If A) then the probability of taking two whites is 1

If B) then the probability of taking two whites is 3/4 * 2/3 = 1/2

If C) then the probability of taking two whites is 1/2 * 1/3 = 1/6

Now, the probability of all balls being white is 2 times as likely as three balls being white and 6 times as likely as 2 balls being white. If all events had equal weight (no bias) then out of 9 possibilities (6 for A + 2 for B + 1 for C), all balls being white has 6 chances.

So, the answer is 6/9 (or 1/3 or 66.67%)

Who can validate this answer???

[Guest]

Concur.

I first went down the road of figuring out the probability of 2 white balls being drawn with each of the following sets of balls WWNN, WWWN, WWWW in the bag. Then I realized that has nothing to do with the probability of the other balls being white that are left. Unless some extra information is given we have to assume there is just as much chance for the remaining balls to be white as non-white. So Dawndada's answer is correct with what information we have.

Dawndada's answer assumes that each outcome from {ww,wn,nn} is equally likely.

P[ww] = P[4WB|2WD]

P[wn] = P[3WB|2WD]

P[nn] = P[2WB|2WD]

Consider all possibilies:

P[4WB] = f(4)

P[3WB] = f(3)

P[2WB] = f(2)

P[1WB] = f(1)

P[0WB] = f(0)

where f(k) is not defined by the OP and must be assumed by the solver.

Let g(k) = P[kWB|2WD] = P[2WD|kWB]*P[kWB]/P[2WD]

= P[2WD|kWB]*P[kWB]/(P[2WD|0WB]*P[0WB] + P[2WD|1WB]*P[1WB] + P[2WD|2WB]*P[2WB] + P[2WD|3WB]*P[3WB] + P[2WD|4WB]*P[4WB])

= P[2WD|kWB]*f(k)/(0*f(0) + 0*f(1) + (1/6)*f(2) + (1/2)*f(3) + 1*f(4))

= 6*P[2WD|kWB]*f(k)/(f(2) + 3*f(3) + 6*f(4))

P[2WD|0WB] = 0

P[2WD|1WB] = 0

P[2WD|2WB] = 1/6

P[2WD|3WB] = 1/2

P[2WD|4WB] = 1

g(0) = 0

g(1) = 0

g(2) = f(2)/(f(2)+3*f(3)+6*f(4))

g(3) = 3*f(3)/(f(2)+3*f(3)+6*f(4))

g(4) = 6*f(4)/(f(2)+3*f(3)+6*f(4))

Now:

P[ww] = g(4)

P[wn] = g(3)

P[nn] = g(2)

Dawndada's and glycereine's answer will only be right if g(4) = g(3) = g(2)

which is true when f(2) = 3*f(3) = 6*f(4)

For symmetry, let f(1) = f(3) and f(0) = f(4)

f(1) = f(3) = 2*f(0)

2*f(0)+ 4*f(0) + 6*f(0) = 12*f(0) = 1

They will be right if

f(0) = 1/12

f(1) = 1/6

f(2) = 1/2

f(3) = 1/6

f(4) = 1/12

This seems to be a fairly arbitrary distribution for white and black balls, and I don't think it would be correct to assume something like this for f(k).

Edited March 22, 2010 by mmiguel1

[Guest]

If A) then the probability of taking two whites is 1

If B) then the probability of taking two whites is 3/4 * 2/3 = 1/2

If C) then the probability of taking two whites is 1/2 * 1/3 = 1/6

Now, the probability of all balls being white is 2 times as likely as three balls being white and 6 times as likely as 2 balls being white. If all events had equal weight (no bias) then out of 9 possibilities (6 for A + 2 for B + 1 for C), all balls being white has 6 chances.

So, the answer is 6/9 (or 1/3 or 66.67%)

Who can validate this answer???

The probability of all balls being white is not 2 times as likely as 3 balls being white.

The way to interpret what you calculated is:

The probability of getting 2 white balls in 2 draws assuming all balls is white is 2 times as likely as the probability of getting 2 white balls in 2 draws assuming 3 balls are white.

The probabilities you calculated say nothing by themselves about how many balls there are in the bag, they both convey information about getting 2 white balls.

The only difference between them is that they assume as fact different numbers of white balls in the bag when trying to convey information about drawing 2 white balls.

So when you divide them you are not dividing probabilities for different number of balls,

you are dividing probabilities of drawing 2 balls with different assumptions.

[Guest]

The probability of all balls being white is not 2 times as likely as 3 balls being white.

The way to interpret what you calculated is:

The probability of getting 2 white balls in 2 draws assuming all balls is white is 2 times as likely as the probability of getting 2 white balls in 2 draws assuming 3 balls are white.

The probabilities you calculated say nothing by themselves about how many balls there are in the bag, they both convey information about getting 2 white balls.

The only difference between them is that they assume as fact different numbers of white balls in the bag when trying to convey information about drawing 2 white balls.

So when you divide them you are not dividing probabilities for different number of balls,

you are dividing probabilities of drawing 2 balls with different assumptions.

--------------

I guess you are right in how I communicated the answer... my point is that given that we didn't get any info other than the results of the random drawing event, I decided we should use that result to figure out the probability of the 4 balls being white (without making an assumption on how the balls were originally drawn). If I have a bag with four balls and ask you to estimate how many balls in the bag are white and I tell you that you can, as a test to use in your answer, draw 2 out of the 4 balls. If you pull 2 balls random and both are white, then I would think that the logic answer is that there is a high chance of all the balls being white (definitely higher than 25%) as we must assume that the result of the test (pulling 2 whites) had a high chance of ocurring.

[Guest]

It seems obvious to me. There is no way to know. We have no information about the other balls.

[superprismatic]

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.

[Guest]

--------------

I guess you are right in how I communicated the answer... my point is that given that we didn't get any info other than the results of the random drawing event, I decided we should use that result to figure out the probability of the 4 balls being white (without making an assumption on how the balls were originally drawn). If I have a bag with four balls and ask you to estimate how many balls in the bag are white and I tell you that you can, as a test to use in your answer, draw 2 out of the 4 balls. If you pull 2 balls random and both are white, then I would think that the logic answer is that there is a high chance of all the balls being white (definitely higher than 25%) as we must assume that the result of the test (pulling 2 whites) had a high chance of ocurring.

If come to any numerical answer at all and you are consistently following the rules of probability, then you are making some sort of assumption on how the balls were initially drawn. This is because the answer directly depends on the probability of getting a bag with 4 white balls given no other information. This probability is the answer to the problem: "You are given a bag with 4 balls. You know nothing else about them other than balls can be black or white. What is the probability that all 4 balls are white?"

You cannot answer this question, because you are not given enough information. In order to answer the original question, you NEED to know the answer to the question I wrote right above, as well as the probability that 3 balls, 2 balls, 1 ball, and 0 balls are white.

To help demonstrate what I mean by the probability that 4 balls are white given no other information, I will give an example.

If I am given a bag and I am told that the probability that all 4 balls are white is 50%, then this is how I should interpret that.

I do not know the contents of my bag. But I know that if I were to replay this game many many times with a different bag each time (always with probability of 4 white balls being 50%), then roughly half of the bags I receive will have 4 white balls, while the other half will have either 0, 1, 2, or 3 white balls in proportions that I cannot calculate with the given information.

This is what probability means: It doesn't make any sure statement about your particular bag, only about the results of opening many many bags in similar situations.

Back to what you said: "If I have a bag with four balls and ask you to estimate how many balls in the bag are white and I tell you that you can, as a test to use in your answer, draw 2 out of the 4 balls."

The first part of this sentence is the same as the problem I talked about above. You cannot estimate how many balls are in the bag without making some sort of assumption about the probability of the initial number of white balls in the bag.

I still think that assuming that when the balls were initially put into the bag, that the probability of a ball being black is the same as the probability of it being white is the best assumption. I must make an assumption in order to move on with your reasoning, otherwise I cannot give you the estimate you asked for, and I choose this assumption.

Here is where the assumption affects everything.

The probability of getting a bag with all black balls is equal to the probability of getting a bag with all white balls.

However, each of these probabilities is less then having some black balls and some white balls.

For example, lets consider having 1 white ball and 3 black balls.

There are 4 ways this can happen.

Each ball is physically distinct and should be treated separately, although the count of the number of white balls does not distinguish between any of the four balls.

Here are the four ways.

WBBB

BWBB

BBWB

BBBW

Compare this to the number of ways of getting only white balls:

WWWW

Each one of these physical outcomes is equally likely according to my assumption.

But 4 of the physical outcomes give 1 white ball and 3 black balls while only 1 physical outcome gives all white balls.

Because the count of white balls does not distinguish between balls, the outcome of getting 1 white ball gets to claim each of the 4 physical states and their corresponding probabilities.

It is thus 4 times more likely to get a bag with 1 white ball and 3 black balls then it is to get a bag with 4 white balls.

Similarly, there are 6 ways of getting 2 white balls.

WWBB

WBWB

WBBW

BWWB

BWBW

BBWW

and getting 2 white balls is 6 times more likely than getting 4 white balls in your bag.

Thus when you draw 2 white balls, you must consider all possible states.

It could be that you have 2 white balls in the bag, and you were lucky enough to draw both of them, or that you drew 2 out of 3 original white balls, or that there were 4 white balls.

Although it is less likely to get 2 white balls if there are only 2 in the bag than it is to get 2 white balls if there are 4 in the bag,

you must also take into account that you are more likely to get a bag with 2 white balls and 2 black balls than you are to get a bag with 4 white balls.

When you take that into account, you get 0.25 for the answer.

If you do not know those probabilities then you are just guessing at the answer.

Edited March 22, 2010 by mmiguel1

[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.

I have the exact same derivation. We only separate at step 4 where you assume that it is equally likely to have 0,1,2,3, or 4 white balls in the bag(0 is as likely as 2 is as likely as 4 etc), while I assume that it is equally likely to have any physical combination of black and white balls in the bag (WWWW is as likely as WWBW is as likely as WBWW is as likely as BWWB etc).

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;

From the wording of the question, I would say we both have valid reasoning and are correct.

From the OP's prior comments, I am guessing you are correct.

[Guest]

If you do not know those probabilities then you are just guessing at the answer.

That wasn't supposed to be there, sorry!

It was supposed to be somewhere above or removed.

[Guest]

Where are you guys getting the notion that the balls must be either black or white? The problem does not state this. It gives no info regarding the balls that are still in the bag.

Edited March 22, 2010 by BasicPoke

[Guest]

Where are you guys getting the notion that the balls must be either black or white. The problem does not state this. It gives no info regarding the balls that are still in the bag.

True. But if we are not allowed to make that assumption then the problem is even more undefined.

You might as well post a question like: "This as a problem. What is the answer?".

That isn't a very fun problem.

It is ok to make assumptions.

For example, 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.

Think about it, we all make assumptions constantly.

Edited March 22, 2010 by mmiguel1