# 4 balls

#### Wilmer

##### In Memoriam
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??

#### MarkFL

Staff member
You've got a lot of balls...

#### Jameson

Staff member
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??
$$\displaystyle P \left[ B | R \mbox{ or } G \right] = \frac{ P \left[ BR \mbox{ or } BG \right]}{P \left[ R \mbox{ or }G \right]}$$

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.

#### Wilmer

##### In Memoriam
$$\displaystyle P \left[ B | R \mbox{ or } G \right] = \frac{ P \left[ BR \mbox{ or } BG \right]}{P \left[ R \mbox{ or }G \right]}$$

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

#### MarkFL

Staff member
Hey there, Denis! (Mooning)

#### Jameson

Staff member
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 $$\displaystyle P \left[ B=1 \vert R=1 \mbox{ or } G=1 \right]$$, where B,R,G are the number of balls of that color. For two draws $$\displaystyle P[B=2]=0$$ 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.

#### Wilmer

##### In Memoriam
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?

#### MarkFL

Staff member
...
@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!

#### Jameson

Staff member
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?

#### MarkFL

Staff member
...
Mark, "Wilmer" is my Godfather's name! Feel better?
Yes, glad to know from where your other nom de forum comes...

#### Wilmer

##### In Memoriam
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:
Probability math problem, thanks!!!!!

#### Jameson

Staff member
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?

#### Wilmer

##### In Memoriam
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?