Boy-Girl probability question
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You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?
probability
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You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?
probability
New contributor
2
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
3
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
3
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
4
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
1
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago
|
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up vote
4
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up vote
4
down vote
favorite
You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?
probability
New contributor
You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?
probability
probability
New contributor
New contributor
edited 4 hours ago
Jam
4,84611431
4,84611431
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asked 5 hours ago
ebramos
342
342
New contributor
New contributor
2
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
3
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
3
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
4
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
1
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago
|
show 4 more comments
2
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
3
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
3
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
4
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
1
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago
2
2
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
3
3
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
3
3
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
4
4
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
1
1
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago
|
show 4 more comments
4 Answers
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The probability is $dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.
The two children correspond to some point $(x,y)$ in $Itimes I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.
The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $dfrac{p-p^2}{2p-p^2}$.
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shamep
can't really be accurately estimated.
– insidesin
2 hours ago
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
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For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.
$frac 12*frac 12$ of all families have two girls. None of them named william.
$frac 12*frac 12=frac 14$ of all families has one girl soaked in skunk urine and a clean boy.
So these families $frac 1m$ of them have the boy named william.
So $frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.
$frac 12*frac 12=frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $frac 1{m}$ have a boy named william.
So $frac 1{4m}$ has skunk urine boy named william and the other a girl.
$frac 12*frac 12 = frac 14$ of all families have two boys.
$frac 1{m^2}$ or $frac 1{4m^2}$ of all families, of these both are named william.
$frac 1{m}frac {m-1}m$ or $frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.
$frac {m-1}mfrac 1{m}$ or $frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.
So $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}= frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.
And $frac 1{4m}+frac 1{4m}=frac 1{2m} = frac {2m}{4m^2}$ of all families have a child named william and the other a girl.
And there are $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}+frac 1{4m}+frac 1{4m} = frac {4m -1}{4m^2}$ of all families has a child named william.
So the probability of a family with two children, one william, having two boys is
$frac {frac {2m-1}{4m^2}}{frac {4m-1}{4m^2}} = frac {2m-1}{4m-1}approx frac 12$ (depending on how rare william is as a name)
and the probability of having a boy and a girl is $frac {2m}{4m-1}approx frac 12$.
....
There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.
The answer is $frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.
But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.
(Let's assume they aren't both named william).
So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $frac 12$.
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These questions are actually a lot of fun. Thank you for posting this!
The essence of this question is really this:
You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.
Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $frac{1}{4}$.
Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.
But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given
that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.
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The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.
$$begin{array}{|c|c|c|}
hline
text{Child 1} & text{Child 2}&text{P(Both Children)}\
hline
BW&BW &p^2\
hline
BW&B'&pleft(frac12-pright)\
hline
BW&G½p\
hline
B'&BW&pleft(frac12-pright)\
hline
B'&B'&left(frac12-pright)^2\
hline
B'&G½left(frac12-pright)\
hline
G&BW½p\
hline
G&B'½left(frac12-pright)\
hline
G&G¼\
hline
end{array}\
$$
In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=frac{P(Bland W)}{P(W)}=frac1pfrac{p^2+pleft(frac12-pright)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=frac12$.
However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.
begin{array}{|c|c|}
hline
text{Child 1} & text{Child 2} &text{P(Both Children)}\
hline
BW&BW&p^2\
hline
BW&B'&left(frac12-pright)p\
hline
BW&G½p\
hline
B'&BW&left(frac12-pright)p\
hline
G&BW½p\
hline
end{array}
So then the probability of the other child being a boy becomes $frac{p^2+left(frac12-pright)p+left(frac12-pright)p}{p^2+left(frac12-pright)p+frac12p+left(frac12-pright)p+frac12p}=frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
add a comment |
4 Answers
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4 Answers
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oldest
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active
oldest
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active
oldest
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up vote
5
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The probability is $dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.
The two children correspond to some point $(x,y)$ in $Itimes I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.
The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $dfrac{p-p^2}{2p-p^2}$.
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shamep
can't really be accurately estimated.
– insidesin
2 hours ago
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
add a comment |
up vote
5
down vote
The probability is $dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.
The two children correspond to some point $(x,y)$ in $Itimes I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.
The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $dfrac{p-p^2}{2p-p^2}$.
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shamep
can't really be accurately estimated.
– insidesin
2 hours ago
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
add a comment |
up vote
5
down vote
up vote
5
down vote
The probability is $dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.
The two children correspond to some point $(x,y)$ in $Itimes I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.
The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $dfrac{p-p^2}{2p-p^2}$.
The probability is $dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.
The two children correspond to some point $(x,y)$ in $Itimes I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.
The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $dfrac{p-p^2}{2p-p^2}$.
edited 3 hours ago
answered 4 hours ago
Steve Kass
11k11429
11k11429
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shamep
can't really be accurately estimated.
– insidesin
2 hours ago
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
add a comment |
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shamep
can't really be accurately estimated.
– insidesin
2 hours ago
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/…
– Steve Kass
3 hours ago
It's a shame
p
can't really be accurately estimated.– insidesin
2 hours ago
It's a shame
p
can't really be accurately estimated.– insidesin
2 hours ago
1
1
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic.
– Steve Kass
2 hours ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
It actually doesn't matter if you allow two Williams in the same family or not. Either way the probability of a family with two children having at least one William is the same and that leads to the same answer. (Suppose a law is brought in forbidding children with the same name and each family with two Williams changes the name of one of them. Every family which had at least one William still does and would answer the the two questions as before, so the probability of a second boy doesn't change.)
– David Hartley
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
@SteveKass This is a good answer and approach but it might be incorrect depending on the OP's true meaning. You've found the probability for when we've drawn a family specified to have a William in it, while the OP's question could equally be asking for drawing a child and then discovering that a child was a William. These are in fact different scenarios and give you different probabilities ($frac12$ or $frac{1-p}{2-p}$). This paper names them C-scenarios and F-scenarios and goes into their differences.
– Jam
1 hour ago
add a comment |
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3
down vote
For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.
$frac 12*frac 12$ of all families have two girls. None of them named william.
$frac 12*frac 12=frac 14$ of all families has one girl soaked in skunk urine and a clean boy.
So these families $frac 1m$ of them have the boy named william.
So $frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.
$frac 12*frac 12=frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $frac 1{m}$ have a boy named william.
So $frac 1{4m}$ has skunk urine boy named william and the other a girl.
$frac 12*frac 12 = frac 14$ of all families have two boys.
$frac 1{m^2}$ or $frac 1{4m^2}$ of all families, of these both are named william.
$frac 1{m}frac {m-1}m$ or $frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.
$frac {m-1}mfrac 1{m}$ or $frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.
So $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}= frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.
And $frac 1{4m}+frac 1{4m}=frac 1{2m} = frac {2m}{4m^2}$ of all families have a child named william and the other a girl.
And there are $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}+frac 1{4m}+frac 1{4m} = frac {4m -1}{4m^2}$ of all families has a child named william.
So the probability of a family with two children, one william, having two boys is
$frac {frac {2m-1}{4m^2}}{frac {4m-1}{4m^2}} = frac {2m-1}{4m-1}approx frac 12$ (depending on how rare william is as a name)
and the probability of having a boy and a girl is $frac {2m}{4m-1}approx frac 12$.
....
There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.
The answer is $frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.
But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.
(Let's assume they aren't both named william).
So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $frac 12$.
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up vote
3
down vote
For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.
$frac 12*frac 12$ of all families have two girls. None of them named william.
$frac 12*frac 12=frac 14$ of all families has one girl soaked in skunk urine and a clean boy.
So these families $frac 1m$ of them have the boy named william.
So $frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.
$frac 12*frac 12=frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $frac 1{m}$ have a boy named william.
So $frac 1{4m}$ has skunk urine boy named william and the other a girl.
$frac 12*frac 12 = frac 14$ of all families have two boys.
$frac 1{m^2}$ or $frac 1{4m^2}$ of all families, of these both are named william.
$frac 1{m}frac {m-1}m$ or $frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.
$frac {m-1}mfrac 1{m}$ or $frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.
So $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}= frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.
And $frac 1{4m}+frac 1{4m}=frac 1{2m} = frac {2m}{4m^2}$ of all families have a child named william and the other a girl.
And there are $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}+frac 1{4m}+frac 1{4m} = frac {4m -1}{4m^2}$ of all families has a child named william.
So the probability of a family with two children, one william, having two boys is
$frac {frac {2m-1}{4m^2}}{frac {4m-1}{4m^2}} = frac {2m-1}{4m-1}approx frac 12$ (depending on how rare william is as a name)
and the probability of having a boy and a girl is $frac {2m}{4m-1}approx frac 12$.
....
There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.
The answer is $frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.
But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.
(Let's assume they aren't both named william).
So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $frac 12$.
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up vote
3
down vote
up vote
3
down vote
For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.
$frac 12*frac 12$ of all families have two girls. None of them named william.
$frac 12*frac 12=frac 14$ of all families has one girl soaked in skunk urine and a clean boy.
So these families $frac 1m$ of them have the boy named william.
So $frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.
$frac 12*frac 12=frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $frac 1{m}$ have a boy named william.
So $frac 1{4m}$ has skunk urine boy named william and the other a girl.
$frac 12*frac 12 = frac 14$ of all families have two boys.
$frac 1{m^2}$ or $frac 1{4m^2}$ of all families, of these both are named william.
$frac 1{m}frac {m-1}m$ or $frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.
$frac {m-1}mfrac 1{m}$ or $frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.
So $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}= frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.
And $frac 1{4m}+frac 1{4m}=frac 1{2m} = frac {2m}{4m^2}$ of all families have a child named william and the other a girl.
And there are $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}+frac 1{4m}+frac 1{4m} = frac {4m -1}{4m^2}$ of all families has a child named william.
So the probability of a family with two children, one william, having two boys is
$frac {frac {2m-1}{4m^2}}{frac {4m-1}{4m^2}} = frac {2m-1}{4m-1}approx frac 12$ (depending on how rare william is as a name)
and the probability of having a boy and a girl is $frac {2m}{4m-1}approx frac 12$.
....
There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.
The answer is $frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.
But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.
(Let's assume they aren't both named william).
So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $frac 12$.
For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.
$frac 12*frac 12$ of all families have two girls. None of them named william.
$frac 12*frac 12=frac 14$ of all families has one girl soaked in skunk urine and a clean boy.
So these families $frac 1m$ of them have the boy named william.
So $frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.
$frac 12*frac 12=frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $frac 1{m}$ have a boy named william.
So $frac 1{4m}$ has skunk urine boy named william and the other a girl.
$frac 12*frac 12 = frac 14$ of all families have two boys.
$frac 1{m^2}$ or $frac 1{4m^2}$ of all families, of these both are named william.
$frac 1{m}frac {m-1}m$ or $frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.
$frac {m-1}mfrac 1{m}$ or $frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.
So $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}= frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.
And $frac 1{4m}+frac 1{4m}=frac 1{2m} = frac {2m}{4m^2}$ of all families have a child named william and the other a girl.
And there are $frac 1{4m^2} + frac {m-1}{4m^2} + frac {m-1}{4m^2}+frac 1{4m}+frac 1{4m} = frac {4m -1}{4m^2}$ of all families has a child named william.
So the probability of a family with two children, one william, having two boys is
$frac {frac {2m-1}{4m^2}}{frac {4m-1}{4m^2}} = frac {2m-1}{4m-1}approx frac 12$ (depending on how rare william is as a name)
and the probability of having a boy and a girl is $frac {2m}{4m-1}approx frac 12$.
....
There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.
The answer is $frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.
But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.
(Let's assume they aren't both named william).
So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $frac 12$.
answered 4 hours ago
fleablood
67.2k22684
67.2k22684
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up vote
0
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These questions are actually a lot of fun. Thank you for posting this!
The essence of this question is really this:
You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.
Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $frac{1}{4}$.
Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.
But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given
that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.
add a comment |
up vote
0
down vote
These questions are actually a lot of fun. Thank you for posting this!
The essence of this question is really this:
You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.
Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $frac{1}{4}$.
Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.
But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given
that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.
add a comment |
up vote
0
down vote
up vote
0
down vote
These questions are actually a lot of fun. Thank you for posting this!
The essence of this question is really this:
You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.
Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $frac{1}{4}$.
Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.
But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given
that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.
These questions are actually a lot of fun. Thank you for posting this!
The essence of this question is really this:
You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.
Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $frac{1}{4}$.
Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.
But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given
that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.
edited 3 hours ago
answered 3 hours ago
Mike
2,694211
2,694211
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The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.
$$begin{array}{|c|c|c|}
hline
text{Child 1} & text{Child 2}&text{P(Both Children)}\
hline
BW&BW &p^2\
hline
BW&B'&pleft(frac12-pright)\
hline
BW&G½p\
hline
B'&BW&pleft(frac12-pright)\
hline
B'&B'&left(frac12-pright)^2\
hline
B'&G½left(frac12-pright)\
hline
G&BW½p\
hline
G&B'½left(frac12-pright)\
hline
G&G¼\
hline
end{array}\
$$
In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=frac{P(Bland W)}{P(W)}=frac1pfrac{p^2+pleft(frac12-pright)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=frac12$.
However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.
begin{array}{|c|c|}
hline
text{Child 1} & text{Child 2} &text{P(Both Children)}\
hline
BW&BW&p^2\
hline
BW&B'&left(frac12-pright)p\
hline
BW&G½p\
hline
B'&BW&left(frac12-pright)p\
hline
G&BW½p\
hline
end{array}
So then the probability of the other child being a boy becomes $frac{p^2+left(frac12-pright)p+left(frac12-pright)p}{p^2+left(frac12-pright)p+frac12p+left(frac12-pright)p+frac12p}=frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
add a comment |
up vote
0
down vote
The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.
$$begin{array}{|c|c|c|}
hline
text{Child 1} & text{Child 2}&text{P(Both Children)}\
hline
BW&BW &p^2\
hline
BW&B'&pleft(frac12-pright)\
hline
BW&G½p\
hline
B'&BW&pleft(frac12-pright)\
hline
B'&B'&left(frac12-pright)^2\
hline
B'&G½left(frac12-pright)\
hline
G&BW½p\
hline
G&B'½left(frac12-pright)\
hline
G&G¼\
hline
end{array}\
$$
In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=frac{P(Bland W)}{P(W)}=frac1pfrac{p^2+pleft(frac12-pright)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=frac12$.
However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.
begin{array}{|c|c|}
hline
text{Child 1} & text{Child 2} &text{P(Both Children)}\
hline
BW&BW&p^2\
hline
BW&B'&left(frac12-pright)p\
hline
BW&G½p\
hline
B'&BW&left(frac12-pright)p\
hline
G&BW½p\
hline
end{array}
So then the probability of the other child being a boy becomes $frac{p^2+left(frac12-pright)p+left(frac12-pright)p}{p^2+left(frac12-pright)p+frac12p+left(frac12-pright)p+frac12p}=frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.
$$begin{array}{|c|c|c|}
hline
text{Child 1} & text{Child 2}&text{P(Both Children)}\
hline
BW&BW &p^2\
hline
BW&B'&pleft(frac12-pright)\
hline
BW&G½p\
hline
B'&BW&pleft(frac12-pright)\
hline
B'&B'&left(frac12-pright)^2\
hline
B'&G½left(frac12-pright)\
hline
G&BW½p\
hline
G&B'½left(frac12-pright)\
hline
G&G¼\
hline
end{array}\
$$
In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=frac{P(Bland W)}{P(W)}=frac1pfrac{p^2+pleft(frac12-pright)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=frac12$.
However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.
begin{array}{|c|c|}
hline
text{Child 1} & text{Child 2} &text{P(Both Children)}\
hline
BW&BW&p^2\
hline
BW&B'&left(frac12-pright)p\
hline
BW&G½p\
hline
B'&BW&left(frac12-pright)p\
hline
G&BW½p\
hline
end{array}
So then the probability of the other child being a boy becomes $frac{p^2+left(frac12-pright)p+left(frac12-pright)p}{p^2+left(frac12-pright)p+frac12p+left(frac12-pright)p+frac12p}=frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.
The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.
$$begin{array}{|c|c|c|}
hline
text{Child 1} & text{Child 2}&text{P(Both Children)}\
hline
BW&BW &p^2\
hline
BW&B'&pleft(frac12-pright)\
hline
BW&G½p\
hline
B'&BW&pleft(frac12-pright)\
hline
B'&B'&left(frac12-pright)^2\
hline
B'&G½left(frac12-pright)\
hline
G&BW½p\
hline
G&B'½left(frac12-pright)\
hline
G&G¼\
hline
end{array}\
$$
In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=frac{P(Bland W)}{P(W)}=frac1pfrac{p^2+pleft(frac12-pright)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=frac12$.
However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.
begin{array}{|c|c|}
hline
text{Child 1} & text{Child 2} &text{P(Both Children)}\
hline
BW&BW&p^2\
hline
BW&B'&left(frac12-pright)p\
hline
BW&G½p\
hline
B'&BW&left(frac12-pright)p\
hline
G&BW½p\
hline
end{array}
So then the probability of the other child being a boy becomes $frac{p^2+left(frac12-pright)p+left(frac12-pright)p}{p^2+left(frac12-pright)p+frac12p+left(frac12-pright)p+frac12p}=frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.
edited 1 hour ago
answered 5 hours ago
Jam
4,84611431
4,84611431
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
add a comment |
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
Argh. I'm mistaken.
– fleablood
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
@fleablood So was I, to be fair.
– Jam
4 hours ago
add a comment |
ebramos is a new contributor. Be nice, and check out our Code of Conduct.
ebramos is a new contributor. Be nice, and check out our Code of Conduct.
ebramos is a new contributor. Be nice, and check out our Code of Conduct.
ebramos is a new contributor. Be nice, and check out our Code of Conduct.
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2
Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck?
– gt6989b
5 hours ago
3
If a couple has two children, what's the probability that both are boys? That exactly one is?
– Ben W
5 hours ago
3
Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which).
– fleablood
5 hours ago
4
@ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small.
– Doug M
4 hours ago
1
Related: math.stackexchange.com/questions/1893041/…
– Henry
2 hours ago