Wonderfully logical illogical probability problem
- Thread starter OSB
- Start date
Ah ok
Seconded
Actually if you say the boy is oldest the probability rises to 100%
Bah!
I don't think it is clear enough mate no. Your query was on the probability of the sex of 2nd child being a boy or girl in both situations. The confusion is arising as folk are thinking that because it is possible in situation 1 that the 2nd child could be older or younger has any bearing on its sex.
I know in probability you need to break down into possible solutions like BG GB etc...but when the question is concerned purely with the sex of the 2nd child, whether it is older or younger can't have any bearing - you can say you have more possible options BG and GB and GG, which you do, but I think it is a mistake to say (in this instance) that because you have one more option (BG) this is relevant to probability of the sex.
But hopefully people can see that is all about ordering. If you have 6 children there are many possible combinations eg:
GGGGGG
GGGGGB
GGGGBG
GGGBGG
GGBGGG
GBGGGG
BGGGGG
GGGGBB
GGGBBG
etc etc
Looking at those top two, if you say ok the first 5 children are girls (as in the top two) the prob of the 6th being a boy is 1/2 (as a boy or a girl birth are equally likely).
However if you only say at least 5 are girls, what is the prob the other one is a boy, the prob is 6/7. This is because there are more ways you can have 5 girls and a boy than there are to have 6 girls.
Is this making sense?
GGGGGG
GGGGGB
GGGGBG
GGGBGG
GGBGGG
GBGGGG
BGGGGG
GGGGBB
GGGBBG
etc etc
Looking at those top two, if you say ok the first 5 children are girls (as in the top two) the prob of the 6th being a boy is 1/2 (as a boy or a girl birth are equally likely).
However if you only say at least 5 are girls, what is the prob the other one is a boy, the prob is 6/7. This is because there are more ways you can have 5 girls and a boy than there are to have 6 girls.
Is this making sense?
Ok i see what you're saying, hopefully my post above with the 6 children shows the situation a little better, i'll do the same for just two.
The 4 possible combinations or boy and girl are:
BB
BG
GB
GG
Now, if we are told that at least one is a girl and then we are asked what is the probability the other is a boy, it is 2/3. Because of the three options containing a girl two also contain a boy. I think this part is clear.
However if we are told that the first child is a girl and then asked what is the probability the other is a boy, it is only 1/2. This is because there are only two options in which the first child is a girl, and then in only one of these is the other child a boy.
Again it is all to do with ordering, combinations and permutations.
Hi mate - yeah I get why you think it is 2/3 - to my mind though BG and GB are the same scenario when the only thing you are concerned with is the sex of the child. How can ordering be relevant to the sex.
I can't get my head to recognise the logic in the fact that just because you have more possible permutations where the child can be a boy than a girl that this should be included as relevant data. If the question was, what is the probability that the 2nd child is a boy and is older say - then ordering becomes relevant?
I can't get my head to recognise the logic in the fact that just because you have more possible permutations where the child can be a boy than a girl that this should be included as relevant data. If the question was, what is the probability that the 2nd child is a boy and is older say - then ordering becomes relevant?
The order is not important, it just takes into account the fact that there are two possible ways for a family with 2 children to have a boy and a girl.
i.e. 1 = Girl, 2 = Boy and 1 = Boy, 2 = Girl
You need to realise the question is about a 2 person set, not a single person, which as mentioned earlier, changes the calculation.
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You say it's not important, do you mean it is not relevant, and so the probability is 50/50 for both situations?
I think what makes this counter intuitive is the phrase:
"What is the probability that his other child is a boy?"
It makes you think that the probability calculation only involves the chances of that child being born male or female. The question isn't just about this singular event. It's about the probability of you encountering a certain type of two-sibling set.
The original question could be re-expressed as:
"Of all two-sibling sets, what is the probability that a given two-sibling set which contains at least one girl, also contains a boy?"
Which I think makes it easier to understand that it's not 0.5.
"What is the probability that his other child is a boy?"
It makes you think that the probability calculation only involves the chances of that child being born male or female. The question isn't just about this singular event. It's about the probability of you encountering a certain type of two-sibling set.
The original question could be re-expressed as:
"Of all two-sibling sets, what is the probability that a given two-sibling set which contains at least one girl, also contains a boy?"
Which I think makes it easier to understand that it's not 0.5.
Are you happy with the answers to the two different statements (poorly worded) in the OP now?
That if we are merely asked "given a girl, what is prob the other a boy" it is 2/3.
But that is we were asked "given the youngest is a girl, what is prob the other is boy" it is 1/2.
It's the same question as in the OP
The information you are being given is:
He has two children (a 2 sibling set)
One of them is a girl [called lisa] (his 2 sibling set contains at least one girl)
What is the chance his 2 sibling set also contains a boy?
Yes, now you know that the only combinations for his children are:
1G then 1B, or
1G then 1G
Both are equally likely. So the chance of the 2nd child being a boy is 1/2 (50%)
Here's where I get confused. I'm not sure that's correct. It's still a two-sibling set. Still the only information we are given is that at least one of the two is a girl. I can't see how that the question is any different to:
"Of all two-sibling sets, what is the probability that a given two-sibling set which contains at least one girl, also contains a boy?"
I believe the answer is only 0.5 in the following case:
"Of all two-sibling sets, what is the probability that a person in a given two-sibling set with at least one girl, is a boy?"
Difference is tho, that there isn't a fixed number of balls.
Just because you've 'removed' a girl from the box doesn't mean there are now less girls and as such a greater chance of a boy.
The annalogy doesn't work.
You seem to have entirely misunderstood the analogy.
All I was saying is that if you pick 2 balls from a total of 2 red and 2 yellow balls they can be arranged in exactly the same combinations as 2 children in a 2 child family.
Not true, you are told two things:
1. At least one is a girl
2. (perhaps more importantly) it is the youngest which is the girl (could have been the oldest though, which it actually is makes no difference, only that one of the two is defined.)
It's different in that this situation (where the girl is the youngest) is the same as saying:
"Of all two-sibling sets, what is the probability that a given two-sibling set in which the first child is a girl, also contains a boy?"
Heh in this case the prob is not 0.5 its 0.67 (2/3) as the set could be GG, GB or BG; giving rise to the two out of three probability.
hope this helps
div0's post clarifies things nicely i believe.
No, you are also told that the girl is his youngest child.
That extra information tells you that there is no chance his first child was a boy.
You now know that he definately had a girl first, then had another child of one sex or the other.
Obviously the chance of that 2nd child being a boy is 50%.
But without knowing that the girl is the oldest or the youngest, then you can't rule out the option of him having 1B, then 1G.
Think of it this way...
If you ONLY know:
1) He has 2 children, and
2) 1 of them is a Girl
Then the possible outcomes are:
First child = B (50%)
First child = G (50%)
Second child = B (50%)
Second child = G (50%)
Chances of him having...
1B first, then 1B second = 0.5*0.5 = 25%
1B first, then 1G second = 0.5*0.5 = 25%
1G first, then 1B second = 0.5*0.5 = 25%
1G first, then 1G second = 0.5*0.5 = 25%
You have met a girl, so you know he hasn't got 2 boys (this is the ONLY thing you know).
So now you know he either had....
1B first, then 1G second (you've met the G)
1G first, then 1B second (you've met the G)
1G first, then 1G second (you've met ONE of the G)
Each of those combinations are equally likely. So there are 2 ways his 'other' child could be a boy (and only 1 way his 'other' child is a girl).
So the chance of his 'other' child being a boy is 2/3.
Now if you are also told that the girl is the YOUNGEST child. Then that leaves the possible options for his family as:
1G first, then 1B second (you've met the G)
1G first, then 1G second (you've met the YOUNGEST G)
Now the chance of his '2nd' child being a boy is 50% (1/2).
The difference is that in the first question you are being asked about the probability of the 'other' child being a boy. The second question is more specific and asks what the probability of his '2nd' child being a boy is, once you know the sex of the first child!
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