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10x(2+5(10-1))/2 = 10x(7x9)/2 = 630/2 = 315 (INCORRECT)
Because you've done it wrong.
10(2+5(10-1))/2 = 10(2+5(9))/2 = 10(2+45)/2 = 470/2 = 235
BODMAS - you have to multiply the 5 by 10-1 [u]before[/u] you add it to the 2
Sock drawer thing - that's probability.
Sex of child - that's statistics.
The sex of the child is probability when expressed as it is in this question. The could replace the child being a boy or a girl with any past event which has a 50/50 outcome:
You ask a woman how many coins she has just tossed and she says two.
Then for some odd reason you ask her "Is at least one a head?", to which she replies "Yes". So what are the odds that she has tossed a head and a tail?
It is still 50:50 for the chances of it being boy and a girl.
You know there is one girl:
So that must surely remove boy/boy and halve the probability of both boy/girl or girl/boy.
100% in my case - boy, boy, boy.
The x must be strong in this one.... 🙂
It is still 50:50 for the chances of it being boy and a girl.You know there is one girl:
So that must surely remove boy/boy and halve the probability of both boy/girl or girl/boy.
Why would it half the probability of boy/girl and girl/boy?
I just tossed a coin twice. One of the times was a head. Does that change the 50/50 chance of getting a head on the other toss?
This thread is illustrating how peer pressure can make people believe errant nonsense.
This thread is illustrating how some people don't understand probability.
😉
It would halve the probability because you cannot have both of them - you know that one is a girl so you get to choose one of boy/girl or girl/boy - not both.
I understand probability perfectly well. Many years of doing engineering maths at uni has seen to that.
It would halve the probability because you cannot have both of them - you know that one is a girl so you get to choose one of boy/girl or girl/boy - not both.
The girl you know about could be the oldest or the youngest, so girl/boy and boy/girl are both still possible outcomes, as is girl/girl. The only outcome you know didn't happen is boy/boy.
I understand probability perfectly well. Many years of doing engineering maths at uni has seen to that.
Any bridges or machines I should keep away from?
I get where the flawed solution comes from, just dont agree that it should be implemented here.
Two coin tosses:
Chance at the beginning that you'll get two of the same side coming up = 50:50.
After the first coin toss chances of you getting a matching one = 50:50.
Why does the child scenario differ?
Its the old gamblers fallicy thing....
Okay Smee answer me this:
[b]with two children what is the probability of mixed sex versus same sex siblings?[/b] *
(*assuming of course that we are still operating in this nice predictable universe where, unlike reality, the probability of a girl or a boy being born is exactly 50%)
[i]Two coin tosses:
Chance at the beginning that you'll get two of the same side coming up = 50:50.
After the first coin toss chances of you getting a matching one = 50:50.[/i]
Your a mile out there.
if you don't know the sex of either then it is 50:50.
Drac - why a mile out?
before first throw your options are:
head/head
tail/tail
head/tail
tail/head
say you get a tail on the first throw your options are now limited to:
tail/tail
tail/head.
and what are the chances of each outcome?
0.25 chance of getting HH
0.25 chance of getting HT
0.25 chance of getting TH
0.25 chance of getting TT
So:
0.25 chance of getting no tails
0.5 chance of getting one tail
0.25 chance of getting two tails
So, you are twice as likely to get one tail as to get two tails.
[img] 
[/img]
[url= http://www.flickr.com/photos/mike_mc/3233045355/ ]http://www.flickr.com/photos/mike_mc/3233045355/[/url]
The other child is either a boy, a girl or a girl…
So, there's 2 in 3 chance of it being a girl.
mike - simply repeating something over and over again doesn't make it correct.
which one of the boy/girl options are you going for on your tree? you cant have them both. your girl must always be a girl she cant change sexes.
mike - read what you've just written.
So the other child is either a boy, a girl or a girl. Surely a girl is the same thing as a girl.
Why can I not have both girl options? Is the girl you know about the eldest child or the youngest?
She is certainly one of them.
[img] 
[/img]
[url] http://www.flickr.com/photos/mike_mc/3233062281/ [/url]
She is certainly one of them
Yes, but she [i]could[/i] be [i]either[/i] of them.
and certainly not the other one.
She could be either, but she has to be one of them - she can't be both.
She does have to be one of them. But there's a [b]probability[/b] that she's either 🙂
If she's the youngest, there's a 50:50 chance the other child is a boy. If she's the eldest, there's a 50:50 chance the other child is a boy.
Just like in my second diagram…
There is a 100% probability that she is not both of them though.
with two children what is the probability of mixed sex versus same sex siblings?
if you don't know the sex of either then it is 50:50.
Correct.
So if you now know that at least one child is a girl then you halve the probability of "same sex" as you know that boy/boy combination is not possible. So "mixed sex" versus "both girls" is 50:25
If she's the youngest that removes the option of her being the oldest and vice versa.
[i]Drac - why a mile out?[/i]
As explained by Mike, it's 2 coins so it's not as simple as H/H T/H T/T
It's
0.25 chance of getting HH
0.25 chance of getting HT
0.25 chance of getting TH
0.25 chance of getting TT
Took me all day to grasp yesterday I did get a little help from a friend also called Graham S who is rather exceptional at maths. He used various ways to explain it but it was the equation on here that made me see the answer.
Drac - go back to sleep.
[i]Drac - go back to sleep. [/i]
That won't make you right.
Ah ha, Smee has just unravelled the mess in my mind, I'm still with the 50% option.
As, if you get a girl first then you have the probabilities of
girl/boy
girl/girl
if you have a girl second then you have
boy/girl
girl/girl
in both scenario's the probability of having a girl and a boy is 50%....as Smee just pointed out you can't stick boy/girl and girl/boy together in this context and the two events are still entirely independent
God damn it I pasted the wrong bit, ok I'm for a nap.
Crikey,
Couldn't help but put my oar in here. To all those clinging on to 50% probability. The final order girl/boy boy/girl is indeed irrelevant the result is counted as the same. However there are two routes to get there. There is only one possible way to get a girl/girl combo. Hence the 2:1 ratio. Those tree diagrams posted earlier explain it all perfectly. Mind you I think it was explained perfectly on about the fourth post so what do I know.
I love these threads. It's better when the people who are wrong tell us how much maths they've done, or how their mum is a midwife so they must be right.
Good God it's catching...
Right.. yes, we all agree that the BIRTHS of these children are entirely independent events where the gender is 50:50 between boy and girl - just like tossing a coin or pulling a sock from an infinite drawer. That's fine.
But in this problem those events have already happened. The woman in the question already has two children and [i]without any extra information[/i] we know that the probabilities are:
all boys = 0.25, all girls = 0.25, mixed = 0.50
Hopefully we are all agreed on that point (if not then read back a bit as it has been shown multiple times)
We are then given additional information that allows us to exclude the all boys possibility, leaving all girls = 0.25 and mixed = 0.50
it also allows you to exclude one of the mixed sex options too. if you disagree tell me why.
Smee, how is that the case?
'At least one girl' accounts for G/B B/G surely? That includes both possible combinations of a mixed sex pair of kiddies?
Isn't that the crux of this whole thing?
So, basically what you're saying is that if you have no previous knowledge about genders but knowing that someone has two kids then the probabilities are
Having a boy and a girl - 50%
Having 2 girls - 25%
Having 2 boys - 25%
But yet if I wasn't to ask how many kids that they have but instead ask what is the gender of your first born (if you have one)
Having a boy - 50%
Having a girl - 50%
I then ask what is the gender of your second child (if you have one)
Having a boy - 50%
Having a girl - 50%
I see where you're coming from but I don't agree, if I have a daughter then there's no reason why my second born is more likely to be a boy
it also allows you to exclude one of the mixed sex options too. if you disagree tell me why.
Because both are still possible.
Wiredchops - you know that one IS a girl, so you cant have both options. It must either be one or the other, it cant be both.