Calculating Probability of Drawing a Blue Ball from an Urn

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Discussion Overview

The discussion revolves around calculating the probability of drawing a blue ball from an urn containing red, green, and blue balls under specific drawing rules. Participants explore different interpretations of the problem, conditional probabilities, and the implications of drawing sequences.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose a conditional probability setup to calculate the likelihood of drawing a blue ball given that at least one of the drawn balls is red or green.
  • There is a discussion about the interpretation of the phrase "if one of the balls drawn is red or green," with some suggesting it could mean the pair contains exactly one blue ball, while others argue it could mean to skip counting if both drawn balls are blue.
  • One participant mentions confirming two different probabilities (5/8 and 3/4) through a program, depending on the interpretation of the conditions.
  • Another participant expresses confusion over the interpretation of the problem and acknowledges a misunderstanding that led to their initial probability calculation.
  • There is a reference to a post by another participant that argues for the 3/4 probability being correct, which some participants agree with, while others still consider the 5/8 probability.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct probability, with multiple competing views remaining regarding the interpretations and calculations involved.

Contextual Notes

Participants express uncertainty about the assumptions underlying their calculations and the implications of different interpretations of the problem statement.

Wilmer
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An urn contaions 1 red ball, 1 green ball and 2 blue balls. An expriment consists of drawing 2 balls in succession from the urn subject to the following rules

i) if the first ball drawn is green, then is is put into the urn before the second ball is drawn
ii) if the first ball drawn is not green, then it is not put back into the urn before the second ball is drawn.

The color of each ball is recorded when it is drawn. If one of the balls drawn is red or green, what is the probability that a blue ball was also drawn??
 
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You've got a lot of balls...
 
Wilmer said:
An urn contaions 1 red ball, 1 green ball and 2 blue balls. An expriment consists of drawing 2 balls in succession from the urn subject to the following rules

i) if the first ball drawn is green, then is is put into the urn before the second ball is drawn
ii) if the first ball drawn is not green, then it is not put back into the urn before the second ball is drawn.

The color of each ball is recorded when it is drawn. If one of the balls drawn is red or green, what is the probability that a blue ball was also drawn??

[math]P \left[ B | R \mbox{ or } G \right] = \frac{ P \left[ BR \mbox{ or } BG \right]}{P \left[ R \mbox{ or }G \right]}[/math]

I believe it can be set up this way. Now it becomes writing out the different ways these things can happen, but I'll wait for confirmation that this is correct so far.
 
Jameson said:
[math]P \left[ B | R \mbox{ or } G \right] = \frac{ P \left[ BR \mbox{ or } BG \right]}{P \left[ R \mbox{ or }G \right]}[/math]

I believe it can be set up this way. Now it becomes writing out the different ways these things can happen, but I'll wait for confirmation that this is correct so far.

Well, I should have said so right off the bat:
I really wanted to see what you took this to mean:
"If one of the balls drawn is red or green,..."

I initially thought it meant: what's the probability
of the pair containing exactly one blue (P=5/8)
but could mean "skip if both blue" (P=3/4).

Hi Mark (Punch)
 
Hey there, Denis! (Mooning)(Angel)
 
Wilmer said:
Well, I should have said so right off the bat:
I really wanted to see what you took this to mean:
"If one of the balls drawn is red or green,..."

I initially thought it meant: what's the probability
of the pair containing exactly one blue (P=5/8)
but could mean "skip if both blue" (P=3/4).

I take it to mean that we assume that 1 of 2 draws is red or green. It could be on the first or second draw, so all of those cases need to be accounted for. Two blues won't be counted obviously, since we need 1 red or green as well.

Put another way the question is [math]P \left[ B=1 \vert R=1 \mbox{ or } G=1 \right] [/math], where B,R,G are the number of balls of that color. For two draws [math]P[B=2]=0[/math] within the conditions of the question.

Did you solve this? What'd you get?

@MarkFL - Sorry to do this, but I had to remove the mooning smilie from our list. I should have done it a while ago but forgot. It's the kind of thing that would upset certain parents about having their child on our site. :(
 
Jameson said:
Did you solve this? What'd you get?
Yes; 5/8 and 3/4 (5/8 if only condition is "exactly one blue").
Confirmed both through a program...

Mark, "Wilmer" is my Godfather's name! Feel better?(Bandit)
 
Jameson said:
...
@MarkFL - Sorry to do this, but I had to remove the mooning smilie from our list. I should have done it a while ago but forgot. It's the kind of thing that would upset certain parents about having their child on our site. :(

I understand completely! :cool:
 
Wilmer said:
Yes; 5/8 and 3/4 (5/8 if only condition is "exactly one blue").
Confirmed both through a program...

If you don't mind showing your work, I'd like to read it to make sure it's the same method. Did you use the conditional probability setup that I posted or something different?
 
  • #10
Wilmer said:
...
Mark, "Wilmer" is my Godfather's name! Feel better?(Bandit)

Yes, glad to know from where your other nom de forum comes...(Dance)
 
  • #11
Jameson said:
If you don't mind showing your work, I'd like to read it to make sure it's the same method. Did you use the conditional probability setup that I posted or something different?

My work is on the 5/8 solution;
the work on the 3/4 solution had already been done;
I confirmed it only.
See both here:
http://www.mymathforum.com/viewtopic.php?f=13&t=33686
 
  • #12
That's the method I proposed, makes sense.

I think the wording is very clear for this kind of problem. I don't see how there could be an interpretation where 2 blue balls are possible. Are you Denis on the other site? If so, it looks like you're writing a computer simulation of the problem.

jks made a great post explaining why 3/4 is correct and not 5/8, which I agree with fully. Do you still have an argument for 5/8?
 
  • #13
Jameson said:
jks made a great post explaining why 3/4 is correct and not 5/8, which I agree with fully. Do you still have an argument for 5/8?
NO! I did explain that my 5/8 was from my WRONG
understanding of the problem: I was solving under
these conditions:
1: draw 1st ball: if Green, put it back
2: draw 2nd ball.
What is probability that exactly one of the pair is Blue?
 

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