Balls in boxes ( probability )

[Rogerio]

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?

[Jin314159]

Is it 1/2?

And shouldn't this be in the Statistics/Set Theory section?

[Rogerio]

No, it's just a brain teaser.
Btw, it seems the correct answer is not 1/2 .

[Jin314159]

Oops. I got 1/2 by doing it in my head. When I did it on paper, I got 2/3. Is that right?

[Rogerio]

...bingo !

[The Bob]

Normally they is an explaination for the stupidier people. :wink: WINK WINK :wink:

The Bob (2004 ©)

[Rogerio]

I would never think of that ! Stupid people doesn't like calculations !

[The Bob]

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 ©)

[NateTG]

\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.

[Gokul43201]

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.

[NateTG]

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.

[BobG]

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.