A couple have 2 children. One of the children is known to be a boy. What is the probability of his sibling being a girl?
Note: it's not a silly trick question.
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Which one is known to be a boy when both are boys { stirs pot maliciously}?
The only info we have is that there are 2 children, one of which is a boy.
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I striggle with these but here i go for broke.
The options are boy or girl. there is a fifty fift chance of any child being a boy or girl. The fact that they already have one boy has no bearing on the chance of the second child as they are unrelated events ( other than they are all related, ba dum tish.........I thankyew)
so my answer
50% chance of being a girl
Now someone smarter than me is going to shoot down my practical logic :(
Off the top-of-me-head: I'd say the probability that the other is a girl is 0.5 minus the probability of the occurrence of identical twins, plus the probability of the occurrence of twins, one of which is the opposite gender.
Wasn't there a long thread about this on the JREF last year (or thereabouts) ?
My money is on "blue bubbles"answer.
however.....
from http://www.sexratio.com/facts.htm.For every 100 baby girls, there are 106 baby boys. This is a fact.
Why more boys? Nobody knows! And it's a real puzzle because...
Other animal species have half male and half female offspring.
Men produce equal numbers of x-chromosome sperms (which make girls) and y-chromosome sperms (which make boys).
I'll allow others to check the sources of this 'amazing fact'
The answer's not 50% ???
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Bugger, I have had a glass or two of the fermented grape and just realised that there are mosaics to considerOriginally Posted by Aardvark
Turners Syndrome and Kleinfelters Syndrome are two that spring to mind
XO
XXY
XYY
Aslo from John's reply post that the answer is not 50%. is this in fact based on the demographics posted by blue bubbles??
51% boy
49% girl
?
Fair enough, John. I take that to mean: it is known of one (unspecified) child that that child is male.
If it is a Monty Hall lookalike ( or Restricted Choice problem, for bridge players), then the answer is 2/3.
Sorry. I've just looked at the question again. On the assumptions, the answer should, of course, be 1/3, not 2/3.
Ah but it's not a Monty Hall lookalike. In the Monty Hall case, you have a priori knowledge that both the car and the goat(s) exist, and can make adjustment accordingly. In John's case, we have no a priori knowledge that a girl child exists.Fair enough, John. I take that to mean: it is known of one (unspecified) child that that child is male.
If it is a Monty Hall lookalike ( or Restricted Choice problem, for bridge players), then the answer is 2/3.
Doing a bit more thinking about the problem...
There are a number of possible answers to this.
Consider the collection of all couples with two children. Now, consider the following two experiments:
For each couple, pick one of the two children at random. If that child is a boy, what is the probability that the other child is a boy? (Answer: the probability that both children are boys is 1 out of 2.)
Note that to perform this experiment, we need to know in advance the sex of only one child.
For each couple, if either child is a boy, what is the probability that the other child is also a boy? (Answer: the probability that both children are boys is 1 out of 3.)
Note that to perform this experiment, we must know in advance the sex of both children.
Begin with a group of 100 families, each with two children, distributed as follows:
25 families with oldest child a boy and youngest child a boy.
25 families with oldest child a boy and youngest child a girl.
25 families with oldest child a girl and youngest child a boy.
25 families with oldest child a girl and youngest child a girl.
Of this group, there are 50 families in which the oldest child is a boy. Of those 50 families, there are 25 families in which the youngest child is also a boy. In other words, out of the group of families in which the oldest child is a boy, 50% have two boys.
From the same group, there are 75 families in which at least one child is a boy. Of those 75 families, there are 25 families in which the other child is also a boy. In other words, out of the group of families in which at least one child is a boy, only 33% have two boys.
This is a bugger of a problem as the wording is ambiguous and it can have several answers.
Someone check my reasoning, please!
They're great these puzzles aren't they?
They key to solving it is recognising that we don't know whether the boy was born first or second; we simply know that out of the 2 children one of them is a boy.
There are 4 ways of having 2 children:
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl
As one is a boy, the girl-girl option is ruled out which leaves:
Boy - Boy
Boy - Girl
Girl - Boy
So if one is a boy there's a 67% chance that the other sibling is a girl.
NOTE: If we know whether the boy was born first or second the chances are 50% as expected.
It's a good little puzzle as it shows how easy it is for intuition to be wrong.
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A lot depends on how we come to know that one child is a boy. Assuming we have asked to see a child at random, I counted the prior possibilities ( seen child first) :
1. B1 B2
2. B2 B1
3. B G
4. G B
5. G1 G2
6. G2 G1
The last three possibilities are ruled out when we see a boy. Of the remainder, twice as many give a boy pair as a mixed pair.
I have checked my answer with Lady Muck, and she is of the firm opinion that I can't count for toffee. And shouldn't be let out on my own... :) She makes it even money.
JJ's answer ( and vbloke's, if I'm following it correctly) is that the question is, in effect: what is the ratio of mixed pairs to boy pairs in the real world? ( 2:1)
So I have come to the conclusion that I don't understand the question :)
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