Brexiteers Destroyed Britains Future,...
News0 min ago
Sex
165 Answers
Following on from the entertaining Dice and Socks threads earlier this week I've now found the problem about the sex of children.
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
Answers
Best Answer
No best answer has yet been selected by factor-fiction. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
If you are now being introduced to the children, FF, you can't answer the question because the information about the probability tree isn't given. Why has he introduced you to Brian first, etc? In that case there would be two events (being introduced to a girl and boy) but all we could say is that
P(next child is a girl) + P(next child is a boy) = 1
which tells us nothing about either probability. Again, this is another common error I have seen, where students seem to think that having found one solution to a problem of this sort with two unknowns, they have found the only solution.
The question asked therefore only makes any sense if you consider the (known) probabilities that Mr. Smith's family contained two boys, or a boy and a girl, or two girls.
P(next child is a girl) + P(next child is a boy) = 1
which tells us nothing about either probability. Again, this is another common error I have seen, where students seem to think that having found one solution to a problem of this sort with two unknowns, they have found the only solution.
The question asked therefore only makes any sense if you consider the (known) probabilities that Mr. Smith's family contained two boys, or a boy and a girl, or two girls.
One in three, factor.
Mr Smith has four possible combinations of gender for his family as has been demonstrated. He can have BB BG GB GG. Since he cannot have GG (because Brian is a boy) that leaves three possibilities of which BB is just one. There are twice as many chances of him having one of each as there are of him having BB.
Mr Smith has four possible combinations of gender for his family as has been demonstrated. He can have BB BG GB GG. Since he cannot have GG (because Brian is a boy) that leaves three possibilities of which BB is just one. There are twice as many chances of him having one of each as there are of him having BB.
factor-fiction: // if the question was rephrased to this:
Mr Smith tells you he has 2 children. He says "look, here's Brian". Given that Brian is a boy what are the chances that the other child is also a boy".
Is that 1/3 or 1/2?
It's 1/3. Try it with six men, each with two children, to each one of which a boy runs up and the man says "Here's [Brian, Tom. etc]." Statistically, the other child of four of these men is a girl, and only of two of them is it a boy.
Just because there are two possible outcomes does not necessarily mean that they're equally likely. This still applies if there are other scenarios in which those two outcomes WOULD be equally likely, and that's what's causing all the confusion here.
Mr Smith tells you he has 2 children. He says "look, here's Brian". Given that Brian is a boy what are the chances that the other child is also a boy".
Is that 1/3 or 1/2?
It's 1/3. Try it with six men, each with two children, to each one of which a boy runs up and the man says "Here's [Brian, Tom. etc]." Statistically, the other child of four of these men is a girl, and only of two of them is it a boy.
Just because there are two possible outcomes does not necessarily mean that they're equally likely. This still applies if there are other scenarios in which those two outcomes WOULD be equally likely, and that's what's causing all the confusion here.
Maybe it's time to wind this thread up. Enough people have been wound up already, but it does show people will see things differently and it's no wonder that if simple cases like this cause confusion, probability theory causes huge confusion in court cases or when analysing medical results for example.
Now to me Monty Hall is a piece of cake compared with this
Now to me Monty Hall is a piece of cake compared with this
The problem is that you are constantly changing the question you are asking. There are three questions here:
"At least one child in a two-child family is a boy. What is the probability that both are boys?" (answer: 1/3).
"The first child in a family is a boy. What is the probability that the next child is a boy?" (Answer, 1/2).
"You are introduced to a boy, one of two children. What is the probability that you are introduced to a boy as the second child?" (Answer: Possibly 1/3, but as you have added two new events in fact the answer is unknown.)
The confusion arises because:
a) People mix the third question with the second, by trying to assume that being introduced to a boy is as likely as being introduced to a girl, when in fact it's not a known probability.
b) People try to introduce the concept of "next" when it doesn't exist (thus confusing the first question with the second).
c) People don't actually sit down and calculate anything, simply listing in their head the probabilities, and then making the false assumption that:
a+b (+c+d+...) = 1 implies a = b (= c = d... ) =1/(number of options)
In every maths problem:
1) decide what the question is.
2) decide what assumptions you are going to make.
3) State them, either when posing the question or when answering it.
Naturally, failing to do any or all of these things will lead to confusions and arguments because people don't have any common ground, and may thus all be "right" in some sense, but right in answering different questions.
"At least one child in a two-child family is a boy. What is the probability that both are boys?" (answer: 1/3).
"The first child in a family is a boy. What is the probability that the next child is a boy?" (Answer, 1/2).
"You are introduced to a boy, one of two children. What is the probability that you are introduced to a boy as the second child?" (Answer: Possibly 1/3, but as you have added two new events in fact the answer is unknown.)
The confusion arises because:
a) People mix the third question with the second, by trying to assume that being introduced to a boy is as likely as being introduced to a girl, when in fact it's not a known probability.
b) People try to introduce the concept of "next" when it doesn't exist (thus confusing the first question with the second).
c) People don't actually sit down and calculate anything, simply listing in their head the probabilities, and then making the false assumption that:
a+b (+c+d+...) = 1 implies a = b (= c = d... ) =1/(number of options)
In every maths problem:
1) decide what the question is.
2) decide what assumptions you are going to make.
3) State them, either when posing the question or when answering it.
Naturally, failing to do any or all of these things will lead to confusions and arguments because people don't have any common ground, and may thus all be "right" in some sense, but right in answering different questions.
"So, NJ, if someone has a boy are you saying the next one is twice aas likely to be a girl as a boy?"
no..No...NO...NOOOOO.
It's not about the liklihood of "the next one", factor. It's about the likely gender combinations of a two child family. It can be two the same (two chances in four), or one of each (two chances in four). Of the "two the same" options only one can be BB, so if you know one child is a boy the GG option can be removed from the possibilities leaving three others. Two of these are "one of each" and one is BB. So it is twice as likely that a couple will have one of each as it is thatthey have two boys.
no..No...NO...NOOOOO.
It's not about the liklihood of "the next one", factor. It's about the likely gender combinations of a two child family. It can be two the same (two chances in four), or one of each (two chances in four). Of the "two the same" options only one can be BB, so if you know one child is a boy the GG option can be removed from the possibilities leaving three others. Two of these are "one of each" and one is BB. So it is twice as likely that a couple will have one of each as it is thatthey have two boys.
In conclusion:
For families with 2.0 children, in which the gender of one child is known, it is a third more likely that both children are of the same gender than in families in which the gender of neither child is known . . . therefore, the other child is twice as likely to be of the opposite gender.
For families with 2.0 children, in which the gender of one child is known, it is a third more likely that both children are of the same gender than in families in which the gender of neither child is known . . . therefore, the other child is twice as likely to be of the opposite gender.
This still confusing some ?
Ok I'm willing to play a game with anyone who has a coin, for say 50 goes.
We'll agree heads represents a girl and tails a boy.
We both toss a coin, if both end up heads we dismiss that as an invalid try, we know that 2 girls are not possible, so we just try again.
If it comes down 2 boys (tails) I shall give the other person £10.
If it comes down 1 boy (tail) and 1 girl (head) they must give me £10.
Fair enough ? 50/50 isn't it ?
Ok I'm willing to play a game with anyone who has a coin, for say 50 goes.
We'll agree heads represents a girl and tails a boy.
We both toss a coin, if both end up heads we dismiss that as an invalid try, we know that 2 girls are not possible, so we just try again.
If it comes down 2 boys (tails) I shall give the other person £10.
If it comes down 1 boy (tail) and 1 girl (head) they must give me £10.
Fair enough ? 50/50 isn't it ?
It's not a ssimple as that though, O-G.
I think we've established that it depends on how we know that "at least one is a boy"
If it's because we picking a family at random from the population of those that have exactly 2 children but not 2 girls then on average it's a 1/3 chance that the other will be a boy.
But if we are just looking at a particular family that has 2 children and we know for example that the first born was a boy because we were the midwife that delivered him then it's a 50-50 (1/2) chance that the second child will be a boy
I think we've established that it depends on how we know that "at least one is a boy"
If it's because we picking a family at random from the population of those that have exactly 2 children but not 2 girls then on average it's a 1/3 chance that the other will be a boy.
But if we are just looking at a particular family that has 2 children and we know for example that the first born was a boy because we were the midwife that delivered him then it's a 50-50 (1/2) chance that the second child will be a boy
I do not know how to convince you but I believe, in fact, it is a very good model of what folk are claiming. IMO I think you are complicating it more than it actually is.
Folk are saying the equivalent of, IF one of us tosses a tail, then is a 50:50 chance whether the other tosses a tail too, or a head.
So we ignore the times when no one tosses a tail, as we know we have seen a boy, even if we do not know which one we have seen. This is much the same as knowing one of us tossed a tail and not caring which.
That leaves just the other possibilities.
Folk are saying the equivalent of, IF one of us tosses a tail, then is a 50:50 chance whether the other tosses a tail too, or a head.
So we ignore the times when no one tosses a tail, as we know we have seen a boy, even if we do not know which one we have seen. This is much the same as knowing one of us tossed a tail and not caring which.
That leaves just the other possibilities.
Latest Posts
Technology1 min ago
Body & Soul16 mins ago
News25 mins ago
Quizzes & Puzzles40 mins ago
