What Is the Probability of Drawing a Second White Ball?

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The discussion revolves around a probability problem involving three indistinguishable boxes, each containing a different pair of colored balls: black+black, black+white, and white+white. After drawing one white ball from a box without seeing its contents, participants debate the probability of drawing a second white ball from the same box. Initial assumptions suggest a probability of 1/2, but further analysis reveals that the correct probability is actually 2/3. This conclusion is reached by considering that if a white ball is drawn, it must have come from either the black+white or white+white box, with the latter being twice as likely to contain another white ball. The conversation highlights the importance of understanding conditional probability and clarifies misconceptions about the problem's parameters.
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You have 3 indistinguishable boxes, containing each one, 2 colored balls: black+black, black+white & white+white.

You open one box and, whithout seeing its interior, you take one white ball.

What is the probability of taking a second white ball from the same box?
 
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Is it 1/2?

And shouldn't this be in the Statistics/Set Theory section?
 
No, it's just a brain teaser.
Btw, it seems the correct answer is not 1/2 .
 
Oops. I got 1/2 by doing it in my head. When I did it on paper, I got 2/3. Is that right?
 
...bingo !
 
Normally they is an explanation for the stupidier people. :wink: WINK WINK :wink:

The Bob (2004 ©)
 
I would never think of that ! Stupid people doesn't like calculations !
 
Rogerio said:
I would never think of that ! Stupid people doesn't like calculations !

Ok the point is I don't understand and I wish to. :biggrin:

The Bob (2004 ©)
 
\frac{2}{3} seems much too high to me.

Consider, the box that is picked must either be B+W or W+W (since it's impossible to pull B+B). Now, barring some kind of sillyness, that leaves a box containing B or a box containing W. Assuming that the boxes were picked with even probability, that's a 50% probability of getting a white ball.
 
  • #10
P(A|B)=P(A^B)/P(B)=(1/3)/(1/2) = 2/3

So, what's wrong with the other argument - Nate's ?

Got it - given that the first pick is W, it's twice as likely to be the WW box as it is to be the BW box.
 
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  • #11
Gokul43201 said:
P(A|B)=P(A^B)/P(B)=(1/3)/(1/2) = 2/3

So, what's wrong with the other argument - Nate's ?

Got it - given that the first pick is W, it's twice as likely to be the WW box as it is to be the BW box.

It's unclear what the process is, so the probability could be anything.

The problem doesn't specify that the white ball is chosen at random. If the problem were something like: You pick one ball from the box, what is the probability that the other ball is the same color? The answer would certainly be 2/3.
 
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  • #12
Rogerio said:
You have 3 indistinguishable boxes, containing each one, 2 colored balls: black+black, black+white & white+white.

You open one box and, whithout seeing its interior, you take one white ball.

What is the probability of taking a second white ball from the same box?








2/3.

There are three white balls you could have pulled out of the box. Of the three, one ball has another black ball in the box. Two balls have another white ball in the box.
 
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