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Valerfore
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simple conditional probability qns
« on: Oct 29th, 2004, 6:05am » |
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http://joshp.typepad.com/15/2003/11/conditional_pro.html
1. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. What is the chance that at least one child is a girl?
Replay with slight change
2. You are at the park, pushing your daughter on a swing. You meet a woman with her daughter and start a conversation. She says she has two children. What is the chance that both children are girls?
Rewind, again
3. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?
my answer is 3/4, 1/2, 1/2, but some people argued that the answer for qn 3 is 1/3.. experts pls help! 
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TimK
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Re: simple conditional probability qns
« Reply #1 on: Oct 29th, 2004, 6:21am » |
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They're right - the answer to question 3 is 1/3. There are three possibilities - GG, GB, BG, and in one of the 3, both children are girls.
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Grimbal
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Re: simple conditional probability qns
« Reply #2 on: Oct 29th, 2004, 8:14am » |
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As I always say, probabilities can not be deduced from a situation, only from a process that brings to that situation. You have to know why the thing that did not happen didn't.
You might assume equal probabilites of having a girl or a boy. But if the woman showed up with a girl, you have to know why not a boy. If she picked a child randomly, there are situations where there was a girl in the family but she showed up with a boy. But these situations are not considered when computing the probability of 2 girls. So, showing up with a girl is not equivalent to answering yes to the question whether she has at least a girl.
If you ask about a girl, you might assume that of the 4 equiprobable cases, 1 of them is eliminated. In that case, all the situations where there is a girl in the family must be considered. The cases are BG, GB and GG, and the probability of 2 girls is 1/3.
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Emul P Edmon
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Re: simple conditional probability qns
« Reply #3 on: Oct 30th, 2004, 11:22am » |
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Right, and the answer to scenario 2 is 1/4 not 1/2. Here are the truth tables
Scenario 1:
B B false
B G true
G B true
G G true
So the odds are 3/4
Scenario 2:
B B false
B G false
G B false
G G true
So the odds are 1/4
Scenario 3:
B G false
G B false
G G true
So the odds are 1/3
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| « Last Edit: Oct 30th, 2004, 11:22am by Emul P Edmon » |
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Sir Col
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Re: simple conditional probability qns
« Reply #4 on: Oct 30th, 2004, 11:40am » |
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Not quite, Emul P Edmon. In the second scenario she arrives with one of her children, which happens to be a girl. Without any further information, and as Grimbal pointed out, we must assume that the gender of the other child is independent. So P(GG)=1/2.
Here's another subtle variation...
4. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She tells you that she has two children and goes on to say, "I am so proud of my eldest, she recently won a poetry competition." What is the chance that both children are girls?
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Grimbal
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Re: simple conditional probability qns
« Reply #5 on: Oct 30th, 2004, 1:25pm » |
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5. You are at the park, pushing your daughter on a swing. You meet a chinese woman and start a conversation. She tells you that she has two children and goes on to say, "I plan to send my eldest daughter to the best university I can get". What is the chance that both children are girls?
PS: just joking.
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rmsgrey
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Re: simple conditional probability qns
« Reply #6 on: Oct 30th, 2004, 1:49pm » |
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It depends for both 4 and 5 on what you think the woman would have said in each of the possible cases.
For 4, if you think the woman picked a girl to talk about, then it's a 1/3 chance of both girls. If you think the choie of which child to mention was independent of gender, then it's a 1/2 chance of both girls.
For 5, you need to judge the probability that the woman would talk about her eldest daughter if she only had one daughter. Thinking about it, you can't assume that the probability she'd talk about her eldest daughter with 2 sons is 0 - she may not have a daughter yet, but may be planning to send the first daughter she subsequently has to a good college.
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mistysakura
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Re: simple conditional probability qns
« Reply #7 on: Nov 6th, 2004, 4:19pm » |
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Lol to number 5. I'm guessing Pr (GG) = 1.
First reason: If there wasn't another daughter, she wouldn't have referred to her as the "eldest" daughter, but just "my daughter".
Second reason: If there was a son, she'd hardly be planning on sending her daughter to the best university, would she? (just kidding.)
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ThudnBlunder
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Re: simple conditional probability qns
« Reply #8 on: Nov 6th, 2004, 8:25pm » |
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Quote:
If there wasn't another daughter, she wouldn't have referred to her as the "eldest" daughter... |
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That's right, if she hadn't been Swiss-Chinese she would have referred to her as the elder daughter. 
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Emul P Edmon
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Re: simple conditional probability qns
« Reply #9 on: Nov 8th, 2004, 4:21pm » |
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Thank you, Sir Col, I had missed that. From now on I'll have to read more carefully. Thanks again...
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ThudnBlunder
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Re: simple conditional probability qns
« Reply #10 on: Nov 10th, 2004, 9:28pm » |
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FRAT HOUSE? : You visit a house wherein a pair of twins live alone. You know that they are either boy/girl fraternal twins or identical girl twins. A girl answers the door. What is the probability that her sibling is also a girl?
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| « Last Edit: Nov 10th, 2004, 9:33pm by ThudnBlunder » |
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Sir Col
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Re: simple conditional probability qns
« Reply #11 on: Nov 11th, 2004, 12:42am » |
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What a delightful variation...
::I shall assume that there is equal chance of the pair of twins being (1) B/G or (2) G1/G2.
There are four combinations of people answering the door, leaving the other person inside:
1B ... G
1G ... B
2G1 ... G2
2G2 ... G1
As a girl answered the door we are dealing with the last three possibilities, each with equal chance. So P(other twin is a girl)=2/3.
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ThudnBlunder
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Re: simple conditional probability qns
« Reply #12 on: Nov 11th, 2004, 4:19am » |
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Quote:| What a delightful variation... |
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But nevertheless varying much more, and somewhat less delightfully, than you have heretofore realized...
Quote:| I shall assume that there is equal chance of the pair of twins being (1) B/G or (2) G1/G2. |
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In this case is that a justified assumption?
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| « Last Edit: Nov 11th, 2004, 10:09am by ThudnBlunder » |
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towr
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Re: simple conditional probability qns
« Reply #13 on: Nov 11th, 2004, 7:47am » |
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on Nov 11th, 2004, 4:19am, THUDandBLUNDER wrote:| In this case is that a justified assumption? |
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Not in general, intuitively fraternal would be more common. But it's hard to estimate without some statistical data on the subject.
Even where they were born can enter into the vacation, seeing as in some places identical twins are unusually common.
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rmsgrey
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Re: simple conditional probability qns
« Reply #14 on: Nov 11th, 2004, 7:58am » |
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For the general version of the frat house twins:
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Assume the twins are identical with probability p, and that, if the twins are fraternal, the boy and the girl have equal chances of opening the door (which may not be justified):
so you have:
(G)B: (1-p)/2
(B)G: (1-p)/2
(G1)G2: p/2
(G2)G1: p/2
giving proability (p/2+p/2) / (p/2+p/2+(1-p)/2) = 2p/p+1 of the other sibling also being a girl
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