Probability all balls are white.

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Homework Help Overview

The problem involves determining the probability that all balls in a bag are white, given that two balls drawn at random are found to be white. The subject area includes probability theory, specifically Bayes' theorem and probability distributions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply Bayes' theorem to calculate the probability but expresses confusion about the assumptions regarding the initial distribution of the balls. Other participants question whether the initial probabilities of the different configurations of balls are equal and discuss the implications of drawing two white balls on the remaining balls.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the assumptions about the initial conditions and the implications of drawing two white balls. Some guidance has been offered regarding the need for prior probabilities, but no consensus has been reached on the correct approach.

Contextual Notes

There is uncertainty about the assumption that all configurations of balls have equal probability, and participants are considering how this affects the calculation of the probability that all balls are white.

Swetasuria
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Homework Statement


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?

2. relevant equations
At school, I'm currently learning Bayes' theorem, probability disribution and Bernoulli trials.

The Attempt at a Solution


The 4 balls can be
1) all white
2) 3 white, 1 non-white
3) 2 white, 2 non-white
4) 1 white, 3 non-white
5) all non-white

P(all white)=1/5

But given, two balls taken at random are white.

I can't figure out what to do next. When I ignore the 2 white balls, I get the answer as 1/3 which is wrong according to my textbook.

[Edit]: I got it!
A-2 balls taken are white
E1-4W
E2-3W, 1Non-white
E3-2W, 2N
E4-1W, 3N
E5-4N

P(E1/A)= {P(E1)P(A/E1)} / {P(E1)P(A/E1)+...+P(E5)P(A/E5)}

={1/5*1} / {1/5*1 + 1/5*3C2/4C2 + 1/5*2C2/4C2 + 1/5*0 + 1/5*0}
=3/5
 
Last edited:
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Swetasuria said:
P(all white)=1/5
Does the problem statement say that all 5 cases have the same probability?
Otherwise, you cannot know this.
 
I'm not too good at this, so may have it wrong, but it seems to me that once you have pulled 2 balls and they are white, what you have is a situation where you have 2 balls (remaining) and you want to know what are the odds that they are both white.

This assumes that the starting condition becomes irrelevant once you draw the first two, and that's where I may be going wrong, though I don't think so.
 
As Bayes showed, the problem is not solvable without plugging in some a priori probabilities. E.g. you could argue that, by and large, bags of balls tended to be, about equally often say, mixed colours or all the same colour. This would be for reasons related to why balls are put in bags in the first place.
Another reasonable choice says each ball is independently white or non-white, but since there are many possible colours these might not be with prob 1/2. Let's say a ball is white with prob p. Now we find that drawing two white balls tells us nothing about the remaining two (as phinds surmised). The prob that they are both white is p2.
 

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