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How about raising a suggestive eyebrow, swivelling one's hips a little, adopting a louche tone of voice, and asking said lady if she'd like any more kids?
What's the probability that she'd say 'yes'?
The order doesn't matter Drac, only the fact that there are twice as many "a boy and a girl" as there are "two girls".
If you were asking "are your children mixed sex or same sex?" then it would be 50/50, but "mixed sex or two girls" is 50/25.
There's always this little one that I heard on the radio the other day:-
A trader enters a bank where everyone else is doing exactly the same trade, and that trade has a 50% chance of making a profit. The trader knows of a trade that has a 75% chance of making a profit.
But in order for the individual trader to get his bonus, two things have to happen: the bank has to make a profit and the individual trader has to make a profit.
So, for the trader to get the best chance of getting his bonus, what should he do?
coffeeking: yeah that one is even subtler.
Basically if you say "Is at least one a girl?" then there is a possible negative response - she could say "No". So when she says "Yes" you can eliminate the No cases.
Whereas if you just ask for the sex of one of them then you don't actually gain any information.
(I think. To be honest that one still makes my head spin a bit)
I came into this and only saw the whole girl/boy, girl/girl thing. I thought I was on a very different thread.
This one will really mess with your minds.
Apparently the sex of one baby can influence the sex of another.
I've read somewhere thatif you have had two children of the same sex, then there is a statistically significant chance that if you have a third, it will be the same sex as the previous two.
[i]The order doesn't matter Drac, only the fact that there are twice as many "a boy and a girl" as there are "two girls".[/i]
I still same that combination as the same girl/boy is no different to boy/girl.
Drac.. order doesn't matter as long as you take into account that it it is twice as likely that two children are boy/girl as it is for them to be girl/girl.
So, if:-
the probability of having girl/girl = x
then
the probability of having boy/girl = 2x
total sum of the probabilies = 3x
therefore
probability of the woman having boy/girl = 2x/3x = 2/3 = 66%
Why the **** didn't you write that in the first place I've got it now. 😳
Mind I never said I was right just couldn't grasp it in writing, maybe I should have wrote an equation down.
Drac.. I know it's hard to believe, what with arguing maths on an internet forum, but lets assume that I have a child X, that child could either be a boy or a girl. Now assume I have another child Y, and that child could be a boy or a girl.
Now, that gives us four combinations:
X B B G G
Y B G B G
As you can see:
the chances of having boy/boy = 25%
the chances of having girl/girl = 25%
Therefore
The chance of having boy/girl (irrespective of order) has to be 50%
If I ever have kids they are gonna hate me for just giving them letters for names though...
😀
See don't write it like that use numbers I then get it. 🙂
Glad you got there Drac. Don't worry, like the Monty Hall problem it is a counter-intuitive head**** for most people.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
The Monty Hall problem I never did get for ages then it sunk in why when watching 21.
What are the odds o nnumber 8 being a certain sex when you were only expecting 7 though??
50%... :o)
Guys the chances of you EVER getting girl/girl are 0%.
I know we all harbour less than discrete fantasies about getting in on a bit of girl/girl lezbo action, but seriously, it's never going to happen.
funkynick
Here's one you should like.
In my pocket, I have a die that I use to play board games with.
What do the sides of the die add up to in total?
miketually - MemberOh, and as for the plane on a conveyor, the wheels on a plane have nothing to do with its ability to take off or not. Air speed over the wings is what matters!
You're correct on both of those. But that still doesn't answer whether or not it takes off. I want to know what, because I'd love it to turn out that you're useless with physics as well as verbal reasoning after your rant about people not understanding probability above [:)]
Your point being?
It was late, I read the question slightly wrong (I read it that the lady had announced the question of the first child, rather than just "one of them"), realised my mistake, then apologised.
Regarding the plane, there are WAY too many variables that were not discussed in the thread to dictate whether or not the plane would take off. Speed of the conveyor belt, wind speed, wind direction, are the engines running or not, required take off speed for the plane etc etc. That's as well as needing to know if the wheel brakes are on or not, and if they're not, the friction coefficient of the conveyor belt, the friction coefficient in the wheel bearings, the drag coefficient of the plane etc etc etc.
Huh? That could be anything. A die is polyhedron, so it depends how many sides it has and how they are numbered surely?
[i]What do the sides of the die add up to in total? [/i]
How many sides?
GrahamS and Drac, fair do's. Most people usually just rush in with "21" as that's what the sides of a "normal" 6 sided die add up to.
There is an equation to work it out though, regardless of the number of sides! Prize to the first person to work it out 😉
Nice try Mboy but I use to play AD&D so was well aware.
[code]n/2 * (n+1)[/code]?
Assuming of course that the die is sequentially numbered from 1.
On the ball GrahamS!
Again, everybody forgets to mention that the die should be sequentially numbered too at this point.
Bonus point for working out the formula for a die that does not necessarily go up sequentially by 1 on each side (though assuming that each increment is equal).
Regarding the plane, there are WAY too many variables that were not discussed in the thread to dictate whether or not the plane would take off. Speed of the conveyor belt, wind speed, wind direction, are the engines running or not, required take off speed for the plane etc etc. That's as well as needing to know if the wheel brakes are on or not, and if they're not, the friction coefficient of the conveyor belt, the friction coefficient in the wheel bearings, the drag coefficient of the plane etc etc etc.
It's a normal plane, taking off as normal on a runway that happens to be moving in the opposite direction to the plane. The runway's not made of glue or treacle and the pilot's not an idiot, so he took the brakes off before trying to take off. The wheel bearings have not been replaced with kryptonite and Superman isn't holding the plan back. Nobody is exceeding the excess baggage allowance and the air traffic controllers are not on strike.
It. Takes. Off.
n(2+d(n-1))/2
n = number of sides
d = increment
Sockpuppet, nearly. Brackets aren't quite all there though.
It's n(2+(d(n-1)))/2
The extra brackets do make the difference.
It is still 50:50. Any 2/3 is flawed maths.
If at least one is a girl you must take the first one as being that girl thereby removing them from the equation. Leaving a 50:50 chance of the next one being a girl.
not sure they do tbh...
does that mean no bonus point? 🙁
Is it just me misreading the brackets or is [i]n(2+d(n-1))/2[/i] not exactly the same as [i]n(2+(d(n-1)))/2[/i]?
And it's 2/3, not 1/2.
In the second you times it by d then add 2 where as in the first you times it by d+2
[teams up with miketually]
yeah!
(elbows owenfracknell out of the way)
[/teams up]
I'm with mike on this one too, there is definitely a redundant set of brackets in there...
The correct answer is C
End of thread.
HTH
Yay smee! Fight the logic. Keep this thread going 🙂
I'm agreeing on it being 50% and I'm also agreeing that those extra brackets are redundant...BIDMAS ordering - brackets, indices, division, multiplication, addition, subtraction
n(2+d(n-1))/2
equates to n(2+dn-d)/2
which is the same as what n(2+(d(n-1)))/2 equates to...n(2+(dn-d))/2
Go on yerself chvck, if you think it's 50% (despite the evidence) then let's see your working.
If i have a drawer with an infinite pair of socks, some blue, some red then if I pull a pair of socks out of the drawer then the probability of being blue first time is 50%, the probability of being blue the second time is 50%. I'm using infinite pairs of socks here as if I pull out a sock from a non-infinite drawer then the number of socks will go down. There is no quantity of boys and girls that exist can be given birth to....it's equal probability and if a girl is the first child then it doesnt reduce the probability of having a girl the next time.
Basically the point that I am trying to labour is that the first birth has absolutely no effect on the second birth meaning that they do not affect the probabilities of each other at all.
The probability of the first birth being a girl was 50% and so the probability of the second birth being a girl is 50%..they are independant probabilities.
I can see how the 66% comes about and it does make perfect sense to me but I do think that boy/girl is effectively the same as girl/boy...you can have a girl and a boy or a boy and a girl, it's the same thing
Anyway I've lost count of the number of times I've repeated myself in this and whether or not any of it is followable and it's too late to care anymore!
Yep socks works too.
Pick two pairs of your socks at random.
Tell me if you have at least one blue pair.
If you do then there is a 66.6% chance that your other pair is red.
Right, using the logic behind the 66% arguement you have to eliminate one of the possible events before calculating the probability
The probability of two independant events is found by multiplying the probabilities together
boy/girl=0.5*0.5=0.25
girl/boy=0.5*0.5=0.25
girl/girl=0.5*0.5=0.25
boy/boy=0.5*0.5=0.25
At this point we can see that girl and boy = 0.25 + 0.25 = 0.5, just because we already know that one is a girl doesn't mean that we can manipulate the probabilities, we can't remove events until after the calculations.
Also - look at it like this if I pick a blue sock then the probability is 50%, then my mate comes along and picks a sock...the colour of the sock that I picked is not going to influence the colour of sock that my mate picks out is it? Surely that's just logical!
note: I do see what you mean and tbh, it's hard for me to judge which is correct as I can actually see them both working depending on how you look at it
Indeed, the events (regardless of socks or child gender) are independent. No one is saying otherwise.
But your own figures contain the answer:
girl and boy = 0.25 + 0.25 = 0.5
girl/girl = 0.25
So a girl and boy is twice as likely as a girl/girl.
OK, I don't actually know anymore, I've royally confused the heck out of myself, I don't know which one is right as they both make sense to me depending on how I look at it - I daresay I could well be wrong. In my brain the logic and the numbers aren't fully agreeing! Either way it's certainly an interesting debate but my face hurts so I'm going to bed!
chvck - MemberOK, I don't actually know anymore, I've royally confused the heck out of myself, I don't know which one is right as they both make sense to me depending on how I look at it - I daresay I could well be wrong. In my brain the logic and the numbers aren't fully agreeing! Either way it's certainly an interesting debate but my face hurts so I'm going to bed!
Easy to do in fairness, but as was pointed out to me originally, we are talking about events that have already happened, not an event that is going to happen (in which case it would be 50/50).
Given that the woman had 2 children, there are 4 possible outcomes, of that we are all agreed.
Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl
It needs to be noted that Boy/Girl is NOT the same as Girl/Boy as these events have already happened. So given as stated, at least one of them is a girl, you can remove the Boy/Boy from the equation and take that there is 2/3 chance of there being a boy and a girl as there were 2 separate opportunities for a combination of the 2 to have happened, whereas there was only 1 opportunity for there to be 2 girls.
Talking about events that are going to happen in the future is totally different though.
Regarding the extra brackets, they DO make a difference.
If we say n=10 and d=5 then if you don't have the extra brackets in n(2+d(n-1))/2 you will get:
10x(2+5(10-1))/2 = 10x(7x9)/2 = 630/2 = 315 (INCORRECT)
Using the extra brackets we get:
10x(2+(5x(10-1)))/2 = 10x(2+45)/2 = 470/2 = 235 (CORRECT)
Sorry to be a pedant 😉
my mum is one of eight kids. she has one brother.
of those eight seven have had children. i've got 23 cousins. only three of us are boys.
of those 23 cousins 4 (all girls) have had children. 4 girls and 2 boys.
christmases used to be fun.
think i best get used to the idea of having daughters.....
Class.
Probability meets statistics again.
Sock drawer thing - that's probability.
Sex of child - that's statistics.
All that shiat on page 2 and most of page 3 of this post, no idea, skipped it.