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Math Question Probability (1 Viewer)

Now you ask anyone with a daughter named Karen, no family will name both their kids Karen, to step f


  • Total voters
    69
I’m pretty bad at working through the reasoning for these types of questions. Voted 1/3 and 1/2. Probably not correct 

 
In question 1, are you asking everyone with 1 daughter to step forward and then from that subpopulation you are asking what are the chances it will be 2 girls?  That would be 1/2.  You already know that sub-population has 1 girl, so the rest would be 50/50 on if the 2nd child was a boy or girl. If you want the total population, then the step forward portion is unnecessary and the answer would be 1/3 for it to be 2 girls (BB, GG, or 1 each).

I applied the same logic to question 2, so I voted 1/2 for both.

This is different from the Monty Hall question, which confuses a lot people.

 
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In question 1, are you asking everyone with 1 daughter to step forward and then from that subpopulation you are asking what are the chances it will be 2 girls?  That would be 1/2.  You already know that sub-population has 1 girl, so the rest would be 50/50 on if the 2nd child was a boy or girl. If you want the total population, then the step forward portion is unnecessary and the answer would be 1/3 for it to be 2 girls (BB, GG, or 1 each).

I applied the same logic to question 2, so I voted 1/2 for both.

This is different from the Monty Hall question, which confuses a lot people.
The sub population has 1 girl so your choice is this

B G

G B

G G

You don't know if the girl is older or younger

 
In question 1, are you asking everyone with 1 daughter to step forward and then from that subpopulation you are asking what are the chances it will be 2 girls?  That would be 1/2.  You already know that sub-population has 1 girl, so the rest would be 50/50 on if the 2nd child was a boy or girl. If you want the total population, then the step forward portion is unnecessary and the answer would be 1/3 for it to be 2 girls (BB, GG, or 1 each).

I applied the same logic to question 2, so I voted 1/2 for both.

This is different from the Monty Hall question, which confuses a lot people.
The sub population has 1 girl so your choice is this

B G

G B

G G

You don't know if the girl is older or younger
I think this is different from the Monty Hall question, but I could be wrong.  If you are starting with a sub population that you know has 2 children, one of which is a girl, you are left with 2 possible scenarios:  GG or GB.  From the phrasing of the question I don't think the age has anything to do with it.

 
you have 100 Dads.

25 x BB ; 25 x GG; 25 x BG; 25 x GB

All the Dads with a daughter step forward which is 75 (25 x GG; 25 x BG; 25 x GB).

In that subgroup you ask for the Dad's with an additional daughter.  That would be the 25 x GG group.  Which is 1/3rd of the subgroup.

 
you have 100 Dads.

25 x BB ; 25 x GG; 25 x BG; 25 x GB

All the Dads with a daughter step forward which is 75 (25 x GG; 25 x BG; 25 x GB).

In that subgroup you ask for the Dad's with an additional daughter.  That would be the 25 x GG group.  Which is 1/3rd of the subgroup.
Ah I can see that.  I do think the question is poorly worded.  If you are only applying it to the subpopulation (which would be 100% with 1 girl) it would be 50%, but as I mentioned in my first response if it is the total population it is 1/3rd.

 
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Ah I can see that.  I do think the question is poorly worded.  If you are only applying it to the subpopulation (which would be 100% with 1 girl) it would be 50%, but as I mentioned in my first response if it is the total population it is 1/3rd.
Yeah a lot of times these questions are hard to parse.  I went 1/3 and 1/2, respectively, because in the first case I think it is the same as asking what percentage of families with at least 1 girl have 2 girls.  Whereas by specifying a gender-specific name you are basically asking what percentage of the girls have a sister.  If you allowed the sisters to have the same names, I believe the second answer would be slightly above 1/2, depending on the popularity of Karen as a girl's name.

 
Yeah a lot of times these questions are hard to parse.  I went 1/3 and 1/2, respectively, because in the first case I think it is the same as asking what percentage of families with at least 1 girl have 2 girls.  Whereas by specifying a gender-specific name you are basically asking what percentage of the girls have a sister.  If you allowed the sisters to have the same names, I believe the second answer would be slightly above 1/2, depending on the popularity of Karen as a girl's name.
And of course they are totally ignoring hermaphrodites :)

 
Seriously, I think B-G and G-B are separate outcomes, so 3/4 that a dad has at least one daughter.  Out of those 1 out of 3 have 2 daugthers.  So 3/4 * 1/3 = 1/4  I do not believe you have the correct answer as an option.

 
The first answer is 1/3.  The second answer is 1/2.  There are a couple of correct explainations for answer 1 already given, so i won't repeat.  

The second question is different, because a family with two girls have twice the chance that one has the name Karen then a family with one girl.  So eventhough only a quarter of the total population has two girls, versus half which have boy and girl, the two girl family are twice as likely to be represented in the selected group.

 
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Seriously, I think B-G and G-B are separate outcomes, so 3/4 that a dad has at least one daughter.  Out of those 1 out of 3 have 2 daugthers.  So 3/4 * 1/3 = 1/4  I do not believe you have the correct answer as an option.
If you stopped one sentence sooner, you had the right answer with the correct rationale..   I have no idea the logic of multiplying by 3/4.  

 
jon_mx said:
If you stopped one sentence sooner, you had the right answer with the correct rationale..   I have no idea the logic of multiplying by 3/4.  
I disagree.  There are 4 possible cases with 2 children: G-G, G-B, B-G, B-B.  3 out of 4 of these possibilities involve having a girl...3/4.  Of that group 1 out of 3 cases involve 2 girls.  Both conditions must be satisfied.  The odds of that are the product of the two  3/4*1/3 

 
I disagree.  There are 4 possible cases with 2 children: G-G, G-B, B-G, B-B.  3 out of 4 of these possibilities involve having a girl...3/4.  Of that group 1 out of 3 cases involve 2 girls.  Both conditions must be satisfied.  The odds of that are the product of the two  3/4*1/3 
The first step is to down select to the three groups with at least ine girl.   After this point, those three types of groups are our population, we no longer include the b-b group.  By multiplying by 3/4 you included them back in and undid the first step.  

 
The first step is to down select to the three groups with at least ine girl.   After this point, those three types of groups are our population, we no longer include the b-b group.  By multiplying by 3/4 you included them back in and undid the first step.  
I think we are interpreting the question differently.  The question asked for the probability that a "randomly selected" dad has 2 daughters.  The original population is the random sample in my eyes, thus both probabilities need to be factored in.  Once you break down to the smaller population, this is no longer a random sample.  It is a sample that already met a certain criteria.

 
I think we are interpreting the question differently.  The question asked for the probability that a "randomly selected" dad has 2 daughters.  The original population is the random sample in my eyes, thus both probabilities need to be factored in.  Once you break down to the smaller population, this is no longer a random sample.  It is a sample that already met a certain criteria.
I agree the wording is not very clear, but the intent was that the random selection was from the group which stepped forward.  Otherwise the question is not interesting. 

 
Never surprises me when folks struggle with conditional probability. 
 

P(A Given B) = P(A and B) / P(B)

 
Ignoring family history since it's not provided. So, assuming no tendencies, just random gender results. I rethought this and decided treat it as a coin flip question. Rewording the question:

"A guy has a coin. He flips it twice. The result from the 1st flip was (fill in the blank). What's chance the 2nd flip is heads?"

If you flipped a coin 0, 1... infinity times. Regardless of the previous flips' results. As an independent event (no twins), the probability of the next flip being heads (or a girl) is still 1/2.

If I'm wrong. That's OK. I've been married a couple of times, so I'm used to it.

 
I'm stumped as to why BG and GB are being treated as separate populations.  The original question doesn't specify order, so I see BG and GB as the same.

 
also is it somewhat depressing that I'm trying to figure out a story problem at 6:30 AM on a Saturday

 
I'm stumped as to why BG and GB are being treated as separate populations.  The original question doesn't specify order, so I see BG and GB as the same.
You could treat them as the same. But there are twice as many of them as there are BB or GG families, so you end up in the same place anyway. Sometimes it's easier to conceptualize this by treating them separately. 

 
Ignoring family history since it's not provided. So, assuming no tendencies, just random gender results. I rethought this and decided treat it as a coin flip question. Rewording the question:

"A guy has a coin. He flips it twice. The result from the 1st flip was (fill in the blank). What's chance the 2nd flip is heads?"

If you flipped a coin 0, 1... infinity times. Regardless of the previous flips' results. As an independent event (no twins), the probability of the next flip being heads (or a girl) is still 1/2.

If I'm wrong. That's OK. I've been married a couple of times, so I'm used to it.
Your scenerio is about the probability of a future random invent which is indeoendant of the past.  That is not the problem here.  We have a population that we assume is random.  We then down select based on some criteria (at least one girl), so we have a different population to choose from.  The population of our group which we then select from would be ecpected to have 50 percent fewer boys in it than girls.  

 
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Say the probability of naming a newborn girl Karen (assuming you don't already have a daughter named Karen) is 0.25.  Then 25% of one-girl families will have a girl named Karen.  0.25 + 0.25(0.75) = 43.75% of two-girl families will have a girl named Karen.  So the answer to part 2 is 0.4375/0.9375 = 46.67%.

:coffee:

 
Say the probability of naming a newborn girl Karen (assuming you don't already have a daughter named Karen) is 0.25.  Then 25% of one-girl families will have a girl named Karen.  0.25 + 0.25(0.75) = 43.75% of two-girl families will have a girl named Karen.  So the answer to part 2 is 0.4375/0.9375 = 46.67%.

:coffee:
No.  Let's say we have 2000 families.  

500 BB

500 GG

500 GB

500 BG

125 of the BG will have a Karen

125 of the GB will have a Karen.

For the GG:  125 will have the first girl named Karen, and 93.75 would have the second girl named Karen (375*.25), total 218.75

Total population left:  468.75

So the answer is definitely 46.67%, not 46.67%.  

Nevermind. 

 
UNTRUE.

I like math. But nobody likes Smoo. So you may be right by chain rule. Or associative principle. Or l'Hopital's rule. Or something.
New probability question

Smoo has 1 post since October 

A new math thread get posted. 

What are the odds he responds within one day? 

At like 4am?

I am so happy to see you. I have some threads for you to read.  PM me for details.

 
Ignoring family history since it's not provided. So, assuming no tendencies, just random gender results. I rethought this and decided treat it as a coin flip question. Rewording the question:

"A guy has a coin. He flips it twice. The result from the 1st flip was (fill in the blank). What's chance the 2nd flip is heads?"

If you flipped a coin 0, 1... infinity times. Regardless of the previous flips' results. As an independent event (no twins), the probability of the next flip being heads (or a girl) is still 1/2.

If I'm wrong. That's OK. I've been married a couple of times, so I'm used to it.
You are wording the question wrong.  If you say atleast one of the flips was heads instead of the first flip was heads, then you get a completely different answer.  You are correct that it is 50/50 if you say the first flip is heads.

But if you say that atleast one of the flips was heads:

A guy flips a coin twice.  Atleast one of the flips is heads.  What are the chances that the other flip was heads?

100 x 1 coin flips.  50 heads and 50 tails. 50 heads reflips and it comes out 25 heads and 25 tails.  50 tails reflip and it comes out 25 heads and 25 tails.

So you end up with 25 x HH; 25 x HT; 25 x TT; 25 x TH.  Now when we ask the question 'Atleast one of the flips was head', you can remove the TT scenario because it does not apply.  You are left with 25 x HH; 25 x HT; 25 x TH. The only solution that satisfies the "what are the chances the 2nd flip is heads' would be the 25 x HH.  The HT and TH do not apply.  So that is 25 out of 75, or 33% chance.

 
What about the likelihood that your second child will have the same sex as the first child?  I think that it's greater than 50%.

 
You have a bag containing N gold coins.  Each night, your neighbor sneaks into your home and steals a coin from your bag (it’s dark, so he can’t see what he’s taking).  He then replaces it with a silver coin.  On the N+1th morning you wake up and randomly pull a coin from the bag.  What is the probability that it’s a gold coin?

 

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