The sub population has 1 girl so your choice is thisIn 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.
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.The sub population has 1 girl so your choice is thisIn 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.
B G
G B
G G
You don't know if the girl is older or younger
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.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.
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.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.
And of course they are totally ignoring hermaphroditesYeah 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.
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.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.
It matters that you down select groups by a specific name.Does it matter if we know the name of the kid?
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*1/3jon_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.
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 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 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.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 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.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.
UNTRUE.Nobody likes math.
Prickly Smoo?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.
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.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.
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.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.
Yep.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.
No. Let's say we have 2000 families.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%.
New probability questionUNTRUE.
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.
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.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.
So of i ask Shakira to sleep with me, it's 50/50!!!!!I'm a firm believer that the true odds of any event occuring is actually 50/50. It either happens, or it doesn't.
I loves me some Smoo.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.