if you cannot have 2 girls....then you have a 50/50 chance of having either 2 boys, or a boy and a girl......because you tell me i automatically have one boy....so that takes the guess work out of one in the series of 2....then the second (not necessarily the eldest or youngest, it wasnt asked) is either a boy or girl...
so
you either have a
B and a girl
or a B and a B
so
you either have a
B and a girl
or a B and a B
50/50....because
you are not defining where the boy is in the series, only that there HAS to be a boy in the series....the only pace in the series that has a variable towards probability is the place in the series vacated by the boy....
if B is in the first place in the series, the only other 2 options for the second place in the series is B or G....
if B occupies the second place in the series, then there are only 2 options for the first place....
50/50
the question doesnt define where the boy is in the series, only that he is in the series.
you are not defining where the boy is in the series, only that there HAS to be a boy in the series....the only pace in the series that has a variable towards probability is the place in the series vacated by the boy....
if B is in the first place in the series, the only other 2 options for the second place in the series is B or G....
if B occupies the second place in the series, then there are only 2 options for the first place....
50/50
the question doesnt define where the boy is in the series, only that he is in the series.
I'm in Austin...
We'd have to narrow our sample set to people that have two kids,
then we'd have to ask the following questions:
Do you have at least one boy? If they said yes, we'd say:
do you have a daughter?
If someone says yes to the first question, chances are 2/3 that
they will say yes to the second.
you are combining the 25% of the random choices that would be BG with the 25% that would be GB, those two combinations as a whole represent 50% of 100%, or 50/75 once you remove the GG combo, or 2/3.
Do this, take two coins, on a piece of paper make 4 columns, HH, TH, HT, TT, flip them at the same time, put a check in the appropriate box, if you get a split result (heads and tails), use the coin on the left as the first in the TH/HT. Do that enough times (say 40) and then report back the percentage of times when you had a tail, you also had a head.
Do this, take two coins, on a piece of paper make 4 columns, HH, TH, HT, TT, flip them at the same time, put a check in the appropriate box, if you get a split result (heads and tails), use the coin on the left as the first in the TH/HT. Do that enough times (say 40) and then report back the percentage of times when you had a tail, you also had a head.
"if B is in the first place in the series, the only other 2 options for the second place in the series is B or G....
if B occupies the second place in the series, then there are only 2 options for the first place...."
True.. but the probability breaks down like this:
BG = 1/3
GB = 1/3
BB = 1/3
Guys, think about logic. It's very, very simple. You don't know the first thing about the other child. Not one thing. Everything about the son is irrelevant. It doesn't apply to the second child.
You just know that there is a second child. That's it. Nothing else to draw upon to come to any conclusions.
50%
You just know that there is a second child. That's it. Nothing else to draw upon to come to any conclusions.
50%
cool I love austin...will plan a trip
ok, deal...if they say yes they have a daughter, and according to your math, they will answer 'daughter' to the second question 66% of the time...every time they answer daughter, then you pay me 300, for every boy that comes up i will give you 200......being that the probability is 2/3 we should be just about even at the evenings close
So... all you people that are still saying 50%, you need to
email the folks at Wikipedia and explain to them that they
got this simple high-school level probability theory problem
wrong.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Second question
Neither order nor age is important. There are four possible child combinations for a two-child family as seen in the sample space above. Three of these families meet the criteria of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:
<table class="wikitable"> <tbody><tr> <th>Older child</th> <th>Younger child</th> </tr> <tr> <td><s>Girl</s></td> <td><s>Girl</s></td> </tr> <tr> <td>Girl</td> <td>Boy</td> </tr> <tr> <td>Boy</td> <td>Girl</td> </tr> <tr> <td>Boy</td> <td>Boy</td> </tr> </tbody></table>
[edit] Bayesian approach
Consider the sample space of 2-child families.
Therefore the probability is 2/3.
email the folks at Wikipedia and explain to them that they
got this simple high-school level probability theory problem
wrong.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Second question
- A random two-child family with at least one boy is chosen. What is the probability that it has a girl?
Neither order nor age is important. There are four possible child combinations for a two-child family as seen in the sample space above. Three of these families meet the criteria of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:
<table class="wikitable"> <tbody><tr> <th>Older child</th> <th>Younger child</th> </tr> <tr> <td><s>Girl</s></td> <td><s>Girl</s></td> </tr> <tr> <td>Girl</td> <td>Boy</td> </tr> <tr> <td>Boy</td> <td>Girl</td> </tr> <tr> <td>Boy</td> <td>Boy</td> </tr> </tbody></table>
[edit] Bayesian approach
Consider the sample space of 2-child families.
- Let X be the event that the family has one boy and one girl.
- Let Y be the event that the family has at least one boy.
- Then:
-
Therefore the probability is 2/3.
