Wonderfully logical illogical probability problem
- Thread starter OSB
- Start date
Correct
What you've said there makes no sense at all! Saying "I have two children, at least one of which is female" is exactly the same as saying "I have two children, this is my youngest child, Lisa"! The information given is the same!!
Age does make a difference.
First situation (no knolwedge about age) "I have two children, this is my daughter Lisa"
We want the prob he has a boy, given he already has a girl, formally noted as P(B|G) which if you look here http://en.wikipedia.org/wiki/Conditional_probability can be re-arrange as the P(B and G)/P(G)
Now from a simple tree diagram this is equal to (1/2)/(3/4) = 2/3
In the second situation (knowledge of age) then this is equivalent to taking a child out of the box (very confusing 'analogy' which isn't actually analogous to this situation!) in that it reduces the probability of the other child to being a boy or a girl to 1/2.
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Yeah of course that's true, i know that if A and B are independent events P(A|B)=P(A). And IF event A was birth of a girl and B the birth of a boy, but that's not what they are. A is of a two child set there is a girl and B of a two child set there is a boy.
Now due the 4 different ways of having girls and boys in a two child set A and B are not independent. The 4 ways are BB, BG, GB and GG.
Therefore as you can see if event A is present (there is a girl) then P(B) is different than were we to just look at P(B)
Doing the maths P(B|A)=2/3
where-as P(B)=3/4
You haven't defined events A and B clearly enough to reflect the situation.
Now due the 4 different ways of having girls and boys in a two child set A and B are not independent. The 4 ways are BB, BG, GB and GG.
Therefore as you can see if event A is present (there is a girl) then P(B) is different than were we to just look at P(B)
Doing the maths P(B|A)=2/3
where-as P(B)=3/4
You haven't defined events A and B clearly enough to reflect the situation.
Ok i'm going to do the problem in a mathematical notation to get rid of all the problems of linguistics and definitions (once they have been dealt with).
I shall define my events as follows:
Event A - the first (therefore the eldest) child (in a set of two) is a girl; P(A)=1/2
Event B - the first child is a boy; P(B)=1/2
Event C - the second child is a girl; P(C)=1/2
Event D - the second child is a boy; P(D)=1/2
Obviously events A and B and events C and D are mutually exclusive.
A tree diagram to help visualise the problem is as follows:
I will use the following notation:
AںB – meaning the union of A and B, that is A or B
A∩B – meaning the intersection of A and B, that is A and B
From the OP, situation one is "I have two children, this is my daughter Lisa" and we want the probability that his other child is a boy. This equates to P((BںD)|((AںC)), solving:
P((BںD)|((AںC)) = P((BںD)∩(AںC)) / P(AںC)
= [P(B∩A)+P(B∩C)+ P(D∩A)+P(D∩C)] / P(AںC)
= [0+¼+¼+0] / ¾
= ½ / ¾
= 2/3
The second situation is "I have two children, this is my youngest child, Lisa". Here the problem is P((BںD)|A), solving:
P((BںD)|A) = P(B∩A)+P(D∩A) / P(A)
= 0+¼ / ½
= 1/2
Hopefully that should be clear enough (though maybe a bit complicated) and should be rigorous enough.
I shall define my events as follows:
Event A - the first (therefore the eldest) child (in a set of two) is a girl; P(A)=1/2
Event B - the first child is a boy; P(B)=1/2
Event C - the second child is a girl; P(C)=1/2
Event D - the second child is a boy; P(D)=1/2
Obviously events A and B and events C and D are mutually exclusive.
A tree diagram to help visualise the problem is as follows:
I will use the following notation:
AںB – meaning the union of A and B, that is A or B
A∩B – meaning the intersection of A and B, that is A and B
From the OP, situation one is "I have two children, this is my daughter Lisa" and we want the probability that his other child is a boy. This equates to P((BںD)|((AںC)), solving:
P((BںD)|((AںC)) = P((BںD)∩(AںC)) / P(AںC)
= [P(B∩A)+P(B∩C)+ P(D∩A)+P(D∩C)] / P(AںC)
= [0+¼+¼+0] / ¾
= ½ / ¾
= 2/3
The second situation is "I have two children, this is my youngest child, Lisa". Here the problem is P((BںD)|A), solving:
P((BںD)|A) = P(B∩A)+P(D∩A) / P(A)
= 0+¼ / ½
= 1/2
Hopefully that should be clear enough (though maybe a bit complicated) and should be rigorous enough.
It applies to both cases, the second case is in essence a single event as div0 explains in the post above. By fixing the age of one child we are no longer looking at possible 2 child combinations but instead just the outcome of one birth. The proof is rigorous and applies to both cases, how can it not.
Yeah seconded, div0's hit the mark (my explanatory skills are somewhat less über!!).
Yeah thats' where i'm at now, another two terms of it at least! I kind of enjoy it really!
Exactly!
I'm doing a Natural Sciences course (Cambridge), so this year i'm doing a module of Maths, one of Physics one Chemistry and one Geology.