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6 girls or 5 girls and a boy are the only 2 scenarios I can think of in that case so that would make it a 50% chance
Wonderfully logical illogical probability problem
- Thread starter OSB
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6 girls or 5 girls and a boy are the only 2 scenarios I can think of in that case so that would make it a 50% chance
6 girls or 5 girls and a boy are the only 2 scenarios I can think of in that case so that would make it a 50% chance
But if you don't know which 5 are girls there are 6 different ways to arrange 5 girls and 1 boy, but only 1 way to arrange 6 girls so 5 girls and 1 boy is 6 times as likely.
Thanks as well, that now makes sense!
Heh no worries, is good for me as well because to be able to explain it one has to really understand it (excuse the posh turn of phrase using 'one' but it removes confusion!). Also people come up with variations on the problem which a good to look at. I have to say i thought it may cause a stir but not this much cafuffle!!
Dont get how it can be anything other than 50/50 given there are only 2 possible outcomes from the question "what is the probablity that the child is a boy or a girl?
The child is a boy
The child is a girl.
Also dont get how "If person A has 9 children all girls what is the probablity of the next child being a boy or a girl".
isnt it still 50/50? Regardless of there being 9 or 99 or 999 girls beforehand. Because in each individual separate child they can only be one or the other.
Thought Jokester was right.
I suck at statistics though
The child is a boy
The child is a girl.
Also dont get how "If person A has 9 children all girls what is the probablity of the next child being a boy or a girl".
isnt it still 50/50? Regardless of there being 9 or 99 or 999 girls beforehand. Because in each individual separate child they can only be one or the other.
Thought Jokester was right.
I suck at statistics though
Dont get how it can be anything other than 50/50 given there are only 2 possible outcomes from the question "what is the probablity that the child is a boy or a girl?
The child is a boy
The child is a girl.
Also dont get how "If person A has 9 children all girls what is the probablity of the next child being a boy or a girl".
isnt it still 50/50? Regardless of there being 9 or 99 or 999 girls beforehand. Because in each individual separate child they can only be one or the other.
Thought Jokester was right.
I suck at statistics though
Agree 100%.
The question isn't about aranging sets of boys and girls, its merely the likelyhood of the remaining child being a boy. Which is 50/50. Wether there were 1 or a million other siblings that were all girls it still doesn't change the chance of the last sibling being a boy. It is an entirely separate entity and has a probability of its own.
Think of the question this way.
A couple has 6 children. All girls.
What is the probability that the next child they conceive is a boy?
Surely that is 50/50, how can it be anything else.
Even though they've had six girls before the chance of a boy next is still 50/50.
Just like flipping a coin, the probabilty of it landing on heads/tails is always 50/50, the previous flips don't effect the outcome of the next one.
Same for the children problem.
That is 50:50. But if you said "If person A has 10 children, at least 9 of which are girls what is the probablity of the remaining child is a boy or a girl", then it wouldn't be.
It's because there are many different ways of having 9 girls and one boy. You could have a boy then 9 girls, a girl then a boy then 8 girls, 2 girls then a boy then 7 girls, etc. Where as there's only one way to have 10 girls. Each situation has an equal probability but there's 10 girls in only one of them.
Dont get how it can be anything other than 50/50 given there are only 2 possible outcomes from the question "what is the probablity that the child is a boy or a girl?
The child is a boy
The child is a girl.
Also dont get how "If person A has 9 children all girls what is the probablity of the next child being a boy or a girl".
isnt it still 50/50? Regardless of there being 9 or 99 or 999 girls beforehand. Because in each individual separate child they can only be one or the other.
Thought Jokester was right.
I suck at statistics though
Taken as an individual case "what is the prob this next child is a boy" the answer is of course 50/50 as whether you have a boy of a girl is not dependent on how many other boys or girls you've had.
However if you look at the situation as a whole (without knowing in which order the children were born) the situation drastically changes. With 9 girls and 1 boy (without the order being known) they could have been born as follows:
GGGGGGGGGB
GGGGGGGGBG
GGGGGGGBGG
GGGGGGBGGG
GGGGGBGGGG
GGGGBGGGGG
GGGBGGGGGG
GGBGGGGGGG
GBGGGGGGGG
BGGGGGGGGG
There there are nine possible ways of having 9Gs and 1B. However there is obv only 1 way of have 10Gs
Therefore if you know that in a group of 10 children that at least 9 are Gs the prob the other child is a B is not 1/2 (50/50) its actually 9/10
You have to be careful of how you word the explainations.
I know what you mean, but by stating it as the '6th' child then people automatically think 'she's had 5 girls, the 6th is either a boy or a girl, so 50% chance'
Like below...
6 girls or 5 girls and a boy are the only 2 scenarios I can think of in that case so that would make it a 50% chance
It's easy to be confused.
You just have to think about the possible outcomes for her family. When you know there are 5 girls, the only possible ways she could have had her children are:
GGGGGG
GGGGGB
GGGGBG
GGGBGG
GGBGGG
GBGGGG
BGGGGG
Obviously there is a far greater chance of the 'other' child being a boy (6 possibilities for the way she had her family), than it being a girl (1 possible way)
Why you naturally think it should be 50% is because you think of her family like:
GGGGGx
GGGGxG
GGGxGG
GGxGGG
GxGGGG
xGGGGG
Where 'x' is 'other' child that you haven't met.
It's easy to think, well in all those cases it could be a boy or a girl and so it doesn't matter, because in all those cases the chance of it being a boy or a girl is 50%.
You're looking at the problem with hindsight and saying she DOES have 5 girls and the 'other' child obviously can only be a boy or a girl, so it's 50/50.
But it doesn't work like that.
You have to look at the possible combinations (and the probabilities) for a 6 person family, THEN apply the information that you know about her children to rule options out. You can't use the information from NOW to try and determine how she HAD her family.
If you get told WHICH of the 5 you have met (ie the youngest 5), then you have determined the order and can rule it down to a simple 50/50 (boy/girl) for the remaining child.
Absolutely - the previous children have no effect on the NEXT child. So in your example you are 100% correct. It is just like flipping a coin, their NEXT child has a 50% chance of being a boy and a 50% chance of being a girl.
Which is why in the OP when you know that the girl is the YOUNGEST, then the answer is 0.5.
But until you know WHICH of the children you are meeting there are more possible ways that the 'other' child is a boy, than a girl.
reading a few posts i think i get it now...managed to force it through my stubborn brain...
in other words:
if i keep rolling a 6 sided dice the greater the chances are that the number 6 will come up.
even though each time i roll the dice its 1/6 that the number 6 will come up.
The effect is cumulative.
If person A had one roll of the dice and person B had ten then theres more chance of the number 6 coming up for person B.
in other words:
if i keep rolling a 6 sided dice the greater the chances are that the number 6 will come up.
even though each time i roll the dice its 1/6 that the number 6 will come up.
The effect is cumulative.
If person A had one roll of the dice and person B had ten then theres more chance of the number 6 coming up for person B.
Correct
Age makes no difference.
Imagine you meet the father and he tells you he has two children and consider they are in a box, at the moment the sample space is:-
BG
GB
BB
GG
B
G
Now imagine we have a new box with two random children in it again and a new father tells you:-
B
G
It's the same again.
Start with another random pair but this time the father tells you
BG
GB
GG
It's different information that he has told you. This is the version that yeilds the 2/3 answer.
In terms of rolling dice, by presenting the daughter is identical to having already rolled that dice for one child.
You can also prove that age makes no difference by doing a probability tree starting with the daughter and working out the odds of younger and older siblings which shows all are equally viable.
Imagine you meet the father and he tells you he has two children and consider they are in a box, at the moment the sample space is:-
BG
GB
BB
GG
Removing a girl from the box leaves only one child in the box, meaning the sample space can only be:-
B
G
Now imagine we have a new box with two random children in it again and a new father tells you:-
Again this removes a girl from the box leaving:-
B
G
It's the same again.
Start with another random pair but this time the father tells you
The sample space is not as above, but because we have only removed the BB option:-
BG
GB
GG
It's different information that he has told you. This is the version that yeilds the 2/3 answer.
In terms of rolling dice, by presenting the daughter is identical to having already rolled that dice for one child.
You can also prove that age makes no difference by doing a probability tree starting with the daughter and working out the odds of younger and older siblings which shows all are equally viable.
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.