Balls in boxes ( probability )

In summary: Therefore, the probability of pulling a second white ball is 2/3. In summary, the problem asks for the probability of pulling a second white ball from a box after already pulling one white ball from the same box. The answer is 2/3. This problem is a brain teaser and not related to statistics or set theory. The answer may seem counterintuitive, but it is derived from the fact that the first white ball could have been pulled from two different boxes, making one box more likely to contain a second white ball.
  • #1
Rogerio
404
1
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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  • #2
Is it 1/2?

And shouldn't this be in the Statistics/Set Theory section?
 
  • #3
No, it's just a brain teaser.
Btw, it seems the correct answer is not 1/2 .
 
  • #4
Oops. I got 1/2 by doing it in my head. When I did it on paper, I got 2/3. Is that right?
 
  • #5
...bingo !
 
  • #6
Normally they is an explanation for the stupidier people. :wink: WINK WINK :wink:

The Bob (2004 ©)
 
  • #7
I would never think of that ! Stupid people doesn't like calculations !
 
  • #8
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 ©)
 
  • #9
[tex]\frac{2}{3}[/tex] 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.
 

1. What is the concept of "Balls in boxes (probability)?"

The concept of "Balls in boxes (probability)" refers to a mathematical problem where a certain number of balls are placed randomly into a certain number of boxes. The probability refers to the likelihood of a specific outcome occurring, such as the number of balls in a particular box.

2. How does one calculate the probability of balls in boxes?

The probability of balls in boxes can be calculated using the formula P = (n!/k!(n-k)!)*(p^k)*(1-p)^(n-k), where n is the total number of balls, k is the number of balls in a specific box, and p is the probability of a ball being placed in that box. This formula is based on the binomial distribution.

3. How does the number of boxes affect the probability in "Balls in boxes (probability)?"

The number of boxes can greatly affect the probability in "Balls in boxes (probability)." As the number of boxes increases, the probability of a ball being placed in a specific box decreases. This is because the total number of outcomes increases, making it less likely for a ball to be placed in a specific box.

4. What is the significance of "Balls in boxes (probability)?"

"Balls in boxes (probability)" is a common problem used in probability and statistics, and is applicable in many real-life scenarios. It helps to understand and calculate the likelihood of a specific outcome occurring in a random distribution.

5. Are there any real-life applications of "Balls in boxes (probability)?"

Yes, there are many real-life applications of "Balls in boxes (probability)." For example, it can be used to calculate the probability of a certain number of defective products in a batch, the chances of winning a lottery, or the likelihood of a disease being present in a population based on a sample size.

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