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34 replies to this topic

### #11 Plumbstar Tom

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Posted 14 June 2011 - 04:52 PM

But Plumbstar, the question is NOT "What is the probability that one of the children is a girl?" It is also not asking, "A mother gives birth to one child, and it is a girl. What is the probability that the second child will be a girl?" The question is "What is the probability that one child is a girl GIVEN THAT the other child is a girl?"

The trick here is to see that both children have already been born. They're both sitting in a room somewhere. The four possible combinations are:

Child 1 boy, Child 2 boy
Child 1 boy, Child 2 girl
Child 1 girl, Child 2 boy
Child 1 girl, Child 2 girl

We KNOW that one of them is a girl, so that immediately eliminates the first scenario. Of the remaining three, only one of them provides ANOTHER girl. That's where the 1/3 probability comes in.

Nope still dont see it

Question states one is a girl whats is the probability that the other is a girl?

Well there are two options it could be a boy or a girl, 1/2

Agreed that this could go on and on though
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### #12 Treesfearme

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Posted 14 June 2011 - 05:50 PM

I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.
We are told that at least one is a girl which removes the (B,b) case.
But (and this is the tricky bit) we have:
(B,G) with the "named" girl being G --> here the other child will be the boy
(G,g) with the "named" girl being G --> here the other child will be the girl "g"
(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!
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### #13 tbrown122387

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Posted 14 June 2011 - 05:50 PM

I second what Shakingdavid said. I heard this problem a while back on a forum that was discussing interview questions for jobs in finance. Both answers are correct depending on the interpretation of the wording.
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### #14 teasemybrain

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Posted 14 June 2011 - 06:50 PM

I disagree. The puzzle's original logic is very sound.
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### #15 superprismatic

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Posted 14 June 2011 - 07:59 PM

It may help to think of the problem in terms of how this could arise in the real world.
Imagine that I am looking at census records. In this census there were two questions
of interest to us: "Do you have two children?" and "Do you have a daughter?". So, we
imagine that we make a pile of records for which both of these questions have a "yes"
answer. Call this pile X. Suppose that another question on the census is "Do you have
a son?" What portion of the records in pile X would you expect to have a "no" answer
to the question about a son?

We can use a coin flip to make fake census records of this type. Each record would
require three coin tosses to make -- one for each of the three questions. Then we can
actually make pile X and count what portion has a "no" response to the question about
a son. I suggest Twinhelix try this and see what happens!
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### #16 Twinhelix

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Posted 14 June 2011 - 08:02 PM

I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.
We are told that at least one is a girl which removes the (B,b) case.
But (and this is the tricky bit) we have:
(B,G) with the "named" girl being G --> here the other child will be the boy
(G,g) with the "named" girl being G --> here the other child will be the girl "g"
(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

Hehe not to be rude but this one made me laugh, cuz at first you say order doesn't matter, the when you get to the (G,g), you made order matter again... I'm sure 2/3 is incorrect?
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### #17 Shay-d

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Posted 14 June 2011 - 08:21 PM

The answer is obvious. The other child is a boy, there is no probability that the child could be a girl because the question clearly states there are two children and ONE is a girl. If there are two children and one girl the other must be a boy.

Then again, if this has to be a probability question (as the question does ask for a probability) 1/3 seems right to me if the order of birth matters, if it does not 50/50 is right because you only have 2 options.

Edited by Shay-d, 14 June 2011 - 08:29 PM.

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### #18 Nebula Prophet

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Posted 14 June 2011 - 08:38 PM

Spoiler for Statistical answer:

Spoiler for Genetic Answer:

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### #19 shakingdavid

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Posted 15 June 2011 - 11:06 AM

I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.
We are told that at least one is a girl which removes the (B,b) case.
But (and this is the tricky bit) we have:
(B,G) with the "named" girl being G --> here the other child will be the boy
(G,g) with the "named" girl being G --> here the other child will be the girl "g"
(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

Not quite right. There is a similar problem for which your solution will work: suppose we have two bowls. One contains a white and a black marble, the other one contains two white marbles. You pick a marble at random, turns out you picked a white marble, what is the probability of the other marble in the same bowl also being white? The answer is 2/3 for this problem as shown by your solution.
But in this case of children, it is in fact twice as likely to have a boy and a girl as it is to have two girls. Consider that no matter what sex your first child is, you have a 1/2 chance to get a boy and a girl (your second child has to be of the other sex). But to get two girls, you must first have a first child girl (probability 1/2), then a second child girl (probability 1/2). The probability for this is 1/2 x 1/2 = 1/4. So if we translate to the marble case, for this to be properly translated, you need two bowls with white+black marbles, and one bowl with white+white. The probability will now be 1/2 for the other marble to be white. Which is the same answer as you get to the boy girl problem with a "named" girl.

So to keep it all straight: 1/3 for the "given that both aren't boys" case, 1/2 for the "given one child who's definitely a girl" case, 2/3 for the marble case which is different but similar.

Edited by shakingdavid, 15 June 2011 - 11:07 AM.

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### #20 phillip1882

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Posted 15 June 2011 - 04:46 PM

to futher comment on super's resoponce, we are obviouly only interested in the case where a person answers yes to the first two questions. this essentially limits the possibilies to being only yes to the third question or no, with 50/50 chances. but let's say i changed the questions to be... do you have two children? are both childern boys? are both children girls? we know the answer is yes to the first question and no to the second question. we can have one coin toss per question, but count an answer of yes, yes, yes to be a miss-toss. now if you ran the simulation, it would be 1/3. the questions however are very similar.
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