Clan x86
General Forums => Entertainment District => Topic started by: Ender on August 14, 2006, 06:34:37 pm
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So a couple has two children. One of them is a girl. What's the probability that the other one is a girl? Explain your reasoning.
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50%. The outcome of something in the past has no affect on the current situation.
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Wrong. ;)
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Vile, odious lies!
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Hehe, I love this problem. Think about it again =)
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Zero.
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No, by "one of them is a girl" I mean at least one. Good thinking though, "one of them is a girl" is sort of unclear.
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25%?
33%.
[I had a change of heart]
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deadly's correct! Good job. Care to explain? (I did say explain your reasoning.)
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There are two kids.
Kid 1 - probably 100% it's a girl
Kid 1 - probably 50% guy / probablity 50% girl
If you combine them there are three options. Guy, Girl, Girl. Since one of them is already a girl, it can't be 66.66%, so it has to be 33.33%. Repeating, of course.
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Once again, good job. I'm going to provide some alternate explanations now. Oh, and btw deadly, real men use fractions.
First, to clarify what deadly said, there are three permutations:
1. guy girl
2. girl guy
3. girl girl
If the other one is a girl then we have two girls, and only one out of three of the possible permutations gives us three girls.
So you may be asking, why isn't there just two options: 1. girl guy 2. girl girl. I said that one of the two children is a girl, not the first one. The first child OR the second child could have been a boy. If you say that the first children was a girl, then there's only two permutations: girl guy and guy girl.
Another way to think about this is that saying "one of them is a girl" is not an event, since you don't know when it happened. Let's say you have three marbles in a bag, two white and one red. If you say you picked out two marbles and one of them was white then you can't determine the probability of you picking out that white marble. If the first one you picked was red, then you have a 100% chance. If the first one you picked was white, then you have a 50% chance. An event requires a timeframe, and you can only apply the rule "prior events don't affect future events" to independent events, and an independent event must be an event in the first place.
You can also do a simulation of this. Keep a variable for the sum of all the outcomes. Let 0 be boy and 1 be girl. Let the sum of the two numbers be the outcome.
0 = two boys
1 = one girl one boy OR one boy one girl
2 = two girls
Generate two numbers in Z & [0, 1] (set of integers intersected with the set of numbers from 0 to 1 inclusive). If both of them are 0's, add one to the sum (since one of them has to be a girl). After you get one outcome add it to the variable that's the sum of all outcomes. You don't need an array for this, an O(n) operation for finding out the fraction of girl-boy/boy-girl combos is inefficient.
So here's the algorithm for finding out the overall result. Let n be your number of trials. If your sum variable == 2n then you got 100% 2-girl permutations. If your sum variable == n then you got 100% mixed permutations. To simplify things, we'll find out the fraction of mixed permutations first. You can find out your fraction of mixed permutations by n / 1 : S / A. S is your sum variable and A is the answer. Thus A = S / n. And the fraction of two-girl combos is 1 - A.
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Or: 0%
Two children, one is a girl. The other is clearly male.
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Two children, one is a girl. The other is clearly male.
It's a probability problem, not a play on words. The statement "one of them is a girl" doesn't necessitate excluding the possibility of the other one being a girl.
Incidentally, this was an easy problem. :(
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Haha, I'll step it up a notch. I have a REALLY REALLY good logic problem that WILL take up a lot of your time. Unfortunately, the book it's in is at home right now, and I'm at work, so I can't post it until I get home. I will do it then though. For now, do you guys know the Monty Hall problem? If you don't, don't google it!
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Once again, good job. I'm going to provide some alternate explanations now. Oh, and btw deadly, real women use fractions.
Fixed it up for you. Besides, 50% was easier to type than 1/2 so that's what I used. :D
Haha, I'll step it up a notch. I have a REALLY REALLY good logic problem that WILL take up a lot of your time. Unfortunately, the book it's in is at home right now, and I'm at work, so I can't post it until I get home. I will do it then though. For now, do you guys know the Monty Hall problem? If you don't, don't google it!
Why would you tell us the name and then tell us NOT to google it? Wouldn't it just be easier to not say anything at all..?
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If I understand correctly, taking a life science approach to this, it's not 33%. I didn't care enough to pay attention back in 6th grade when we talked about sex (I lit my binder on fire using a magnifing glass one day, though. Teacher yelled at me), but I think there's a chance somewhat less than 50% for a girl to be born, and a bit more than 50% for a guy, and that it doesn't change at all after the first child.
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If I understand correctly, taking a life science approach to this, it's not 33%. I didn't care enough to pay attention back in 6th grade when we talked about sex (I lit my binder on fire using a magnifing glass one day, though. Teacher yelled at me), but I think there's a chance somewhat less than 50% for a girl to be born, and a bit more than 50% for a guy, and that it doesn't change at all after the first child.
It's a probability problem. When presented with a situation like that, you give them equal odds.
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Why, Joe, are you expecting?
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Yes, you know I definately got laid. :p
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You probably just sucked a double-headed dildo on a hovercraft.
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You probably just sucked a double-headed dildo on a hovercraft.
R O F L.