You have a bag with a ball in it.
The ball is either black or white. (equal probabilties)
You put a white ball in the bag and shake it up.
You pull out a white ball but can't tell if it's one you just put in or was the original one.
What is the probability that the ball in the bag is white?
2/3
There are 2 balls, each can be either white or black.
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
P(A|B) = P (A and B)/P(B) => result.
Although conditional probability takes a little bit of the magic away from these puzzles. Here is a decent article if you'd like to be able to do more puzzles like this.
Although conditional probability takes a little bit of the magic away from these puzzles. Here is a decent article if you'd like to be able to do more puzzles like this.
QUOTE (NeoDevin+Jun 2 2008, 11:51 PM)
There are 2 balls, each can be either white or black.
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
But the second "white ball in the bag" has a black ball in the hand, and is thus cannot occur, leaving only two possible cases and a 1:2 probability.
Merely playing Devil's advocate...
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
But the second "white ball in the bag" has a black ball in the hand, and is thus cannot occur, leaving only two possible cases and a 1:2 probability.
Merely playing Devil's advocate...
Oh man - this reminds me so much of a probability riddle thread that went totally out of hand here. It's a perfect example of how people can start assuming information in their heads about a given problem and relentlessly defend an incorrect answer because of it.
Not that that happens here. ¬_¬
Fortunately, this question gives sufficient information to yield an answer of 2/3.
Not that that happens here. ¬_¬
Fortunately, this question gives sufficient information to yield an answer of 2/3.
QUOTE (NeoDevin+Jun 2 2008, 11:51 PM)
There are 2 balls, each can be either white or black.
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
A better illustration might be:
Let Ball A be the ball in the bag, and Ball B be the ball we add. The possible outcomes are thus:
Take out Ball A, it is black, Ball B is in the bag and is white.
Take out Ball A, it is white, Ball B is in the bag and is white.
Take out Ball B, it is white, Ball A is in the bag and is black.
Take out Ball B, it is white, Ball A is in the bag and is white.
Since the first one cannot happen (we know the ball we took out is white), there are three remaining possibilities. In 2 of those cases, the ball in the bag is white. Thus, the correct answer is 2/3.
This gives 4 possibilities:
Both are black
Both are white
The one in hand is black, the one in the bag is white
The one in hand is white, the one in the bag is black
Because we know that at least one is white, the first case cannot occur. Of the other three cases, 2 have a white ball in the bag. Therefore the answer of 2/3 is correct.
A better illustration might be:
Let Ball A be the ball in the bag, and Ball B be the ball we add. The possible outcomes are thus:
Take out Ball A, it is black, Ball B is in the bag and is white.
Take out Ball A, it is white, Ball B is in the bag and is white.
Take out Ball B, it is white, Ball A is in the bag and is black.
Take out Ball B, it is white, Ball A is in the bag and is white.
Since the first one cannot happen (we know the ball we took out is white), there are three remaining possibilities. In 2 of those cases, the ball in the bag is white. Thus, the correct answer is 2/3.
QUOTE (LostInThought+Jun 3 2008, 12:23 PM)
Oh man - this reminds me so much of a probability riddle thread that went totally out of hand here. It's a perfect example of how people can start assuming information in their heads about a given problem and relentlessly defend an incorrect answer because of it.
Not that that happens here. ¬_¬
Fortunately, this question gives sufficient information to yield an answer of 2/3.
Since we're all so fond of controversy
, I'm posting the riddle that started that thread:
I know how I'd answer.
Not that that happens here. ¬_¬
Fortunately, this question gives sufficient information to yield an answer of 2/3.
Since we're all so fond of controversy
, I'm posting the riddle that started that thread:QUOTE
You get new neighbours, a family with two kids. If you see a boy standing at a window; what is the probability of the other kid being a girl?
So your answer will probably take the form of a fraction. Pretend that the chance of any given kid being a boy is exactly 1/2 and the chance of being a girl is exactly 1/2.
So your answer will probably take the form of a fraction. Pretend that the chance of any given kid being a boy is exactly 1/2 and the chance of being a girl is exactly 1/2.
I know how I'd answer.
Dont get the same answer.
"Since the first one cannot happen (we know the ball we took out is white),
there are three remaining possibilities."
(you don't know which ball you drew)
"There are 2 balls, each can be either white or black.
(only the original)
buttershug:
probability of remaining ball in bag being white,
after pulling a white ball from bag is s.
b = p(black), w = p(white)
b1 in the bag with w = 1/2 and b = 1/2
b2 put in bag with w = 1 and b = 0
p(b1)=p(b2)=1/2
only 3 possibilities:
pull b1 in b, s = 1
pull b1 in w, s = 1
(mutually exclusive, one or the other)
pull b2 in w, s = 1/2 (b1 is still unknown)
s = 1/2(1+1)+1/2(1/2) = 3/4
"Since the first one cannot happen (we know the ball we took out is white),
there are three remaining possibilities."
(you don't know which ball you drew)
"There are 2 balls, each can be either white or black.
(only the original)
buttershug:
QUOTE
The ball is either black or white. (equal probabilties)
probability of remaining ball in bag being white,
after pulling a white ball from bag is s.
b = p(black), w = p(white)
b1 in the bag with w = 1/2 and b = 1/2
b2 put in bag with w = 1 and b = 0
p(b1)=p(b2)=1/2
only 3 possibilities:
pull b1 in b, s = 1
pull b1 in w, s = 1
(mutually exclusive, one or the other)
pull b2 in w, s = 1/2 (b1 is still unknown)
s = 1/2(1+1)+1/2(1/2) = 3/4
With morning comes the dawn!
I'm not answering the same question!
Another reason why I don't work for NASA.
Carry on.
I'm not answering the same question!
Another reason why I don't work for NASA.
Carry on.
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