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/sci/ - Science & Math

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3088061 No.3088061 [Reply] [Original]

Let's see if /sci/ can answer this probability question...
John Proctor has two children. Atleast one of them is a girl. What is the probability of both children being girls?

Don't dissapoint me /sci/

>> No.3088066

50%

duh

>> No.3088068

1/3

>> No.3088075

If he's only having 2 children and it's given that 1 of them is a girl, you're left with one other child... And since there are only 2 sexes of children... There's a 50% chance of another girl being born, leaving you with two girls! Probably posting in a troll thread.

>> No.3088076

>>3088066
>>3088068
Prove it

>> No.3088079

>>3088068

To expound on that, the possibilities are:

BB
BG
GB
GG

It can't be BB because atleast one is a girl. Hence 1/3.

>> No.3088080

>>3088075
You are trolling in a troll thread.
The answer is 1/3, OP.

>> No.3088090

Hey guys. The answer is actually 1/3, not 1/2 as you would expect

The possibilities for 2 kids are:

bb, bg, gb, gg b = boy, g = girl

if you know at least one is a girl then this narrows it down to:

bg, gb, gg

so gg is 1/3

Its only 1 in 2 if you know that a particular child is a girl, say the eldest, but if you only know at east 1 is then its 1/3

>> No.3088091

Girl/Sage
Sage/Girl
0%

>> No.3088101

50% unless you think sex chromosomes undergo meiotic drive

>> No.3088132

1/4 50% chance for a 1 girl. Then 50% chance for another
so you half the chance for the first child again. I feel this is right but my mind works in strange ways so i cant explain it properly.

>> No.3088139

>>3088079

BG and GB are the same thing....

that leaves

GG
GB/BG

50%

>> No.3088143

>>3088090
Knowing that the other is a girl is irrelevant, there is always a chance that the child is a boy aswell.

>> No.3088152

>>3088061
25%

It doesn't matter what the first one is, as 25% is the probability that they're both girls.

>> No.3088155

>>3088132
simple XOR operation

>> No.3088161

>>3088139
>implying permutations are combinations

>> No.3088177

>>3088152
You're assuming that there's a 25% chance they're both boys, which there's not.

>> No.3088180

>>3088161

This is a permutation?

Genuine question, also I'm not the guy you responded to.

>> No.3088189

>>3088177
No, I'm just answering this question
>what is the probability of both children being girls?

Even if you know one is a girl, that doesn't change the probability answer to the question.

>> No.3088195

*sigh*
if the question was: a specific one is a girl, whats the chances of the other being a girl
>50%

if the question was: at least one is a girl, whats the chances of there being girls
>100%

if the question was: whats the chances of both being girls
>25%

if the question was: whats the chances of at least one girl
>75%

the question IS: at least one is a girl, whats the chances of both being girls
>33.33%

>> No.3088204

>>3088189
25% chance boy, girl
25% chance girl, boy
25% chance girl, girl
???

>> No.3088206

>>3088189
it absolutely does. because we know that having two boys ISNT an option, there are only 3 permutations possible, of which one is two girls

>> No.3088209

>>3088139

GB/BG is twice as likely as GG though.

Rearranging the possibilites won't stop the answer from being 1/3.

This is a well known paradox, called the two-child problem. Similiar to the Monty-hall problem (The thing with the three doors and a goat) It seems like the answer is 1/2, but actually its 1/2

>> No.3088212

People who answer 50% or 25% make me want to fucking kill myself.

>> No.3088218 [DELETED] 

All i'm saying is, the OP never specified if it was child # 1 or child # 2 therefore it's random and GB and BG will be the say thing.

theres either 1 boy and 1 girl or 2 girls.

2 possibilities. 50%

>> No.3088220

>>3088204
no
25% bg
25% bb
25% gb
25% gg

>what is the probability of both children being girls?

>> No.3088221

people that answer 33%, why does order matter for BG/GB but not for GG/GG, just because the variables are different doesn't mean you change the entire concept

at least one girl, the possibilities are:

1st boy, 2nd girl
1st girl, 2nd boy
1st girl, 2nd girl
1st girl, 2nd girl

that's 50%

>> No.3088226

>>3088189
hey you. i have a proposition. there are three doors here, and look:
>opens door number 1
its a years supply of oreos! however, behind doors 2 and three there's a brand new free luxury car and a bag of old fertilizer. you dont know which one is behind which door, though.

so here's the deal. you can either STAY and keep the oreos, or SWITCH and pick either door 2 or 3, trying to get that car. what do you say, switch or stay?

>> No.3088234

All i'm saying is the OP never specified if it was child #1 or child #2 therefore it's random and GB and BG are the same thing.

there's either:

1 girl and 1 boy
or
2 girls

50%

/thread

>> No.3088238

>>3088226
i mean come on, if you choose to switch, there's a 50% shot of getting that car right?

>> No.3088240

>>3088220
No, it can't be two boys. You're an idiot.

>> No.3088246

>>3088209
>but actually its 1/2

obvs meant 1/3

>> No.3088250

>>3088234
you forgot one boy and one girl. just because together they add up to one of each doesnt make it the same thing. thats like saying "hey kids at least one of you is getting allowance this week, what are the odds of both of you getting allowance?" and then thinking that them getting it and you not, is the same as vice versa.

>> No.3088264

>>3088250

so even if i take that into account... then you also have to take into account what the guy above said:

you can have:
girl boy
boy girl
girl girl
girl girl

still 50%

>> No.3088282

>>3088061
From my point of view? 50%

I have no idea who John Proctor is or why he was picked. I am told that one of his children is a girl, and the question at hand is "What is the probability of the other child also being a girl?"

My only answer can be "50%".

From your point of view, it is possible that you picked John Proctor randomly from the set of parents with two children, and at least one girl. In that case, your additional information would allow you to say that the probability of both children being girls is 1 in 3.

On the other hand, if you met John Proctor at random, and he mentioned spontaneously that he had a daughter, and also that he had two children, you would be as ignorant as I am and could only say the probability is 50%.

If he told you that he has two children, both girls, and introduced them, then to you the probability is 100%.

I have no information on how you selected John Proctor or gathered the given information, I only know there is one child whose sex I have no information on.

Probability is not an absolute quantity, but a statement of how much confidence you can have in any given prediction based on the information available.

>> No.3088286

>>3088240
so you're saying that some how having a girl makes the probability of another girl less likely?

>> No.3088293

>>3088240
There is a non-zero probability that OP is lying, and John Proctor has two boys.

There is also a non-zero probability that OP is lying, and there is no John Proctor.

>> No.3088294

>>3088286
How the fuck did you get that from my post?

>>3088061
>Atleast one of them is a girl.

So BB is not an option.

>> No.3088316

Possible combinations:

BOY 1 BOY 2
BOY 2 BOY 1
BOY1 GIRL1
GIRL1 BOY1
GIRL1 GIRL2
GIRL2 GIRL1

if we eliminate 2 boys (impossible) we are left with:

BOY1 GIRL1
GIRL1 BOY1
GIRL1 GIRL2
GIRL2 GIRL1

50%

>> No.3088321
File: 245 KB, 3750x3750, 1303279485138.jpg [View same] [iqdb] [saucenao] [google]
3088321

order is insignificant
so instead of going based on permutations {GB, BG, GG}
we use instead combinations {GB, GG}
which looks pretty fucking 50% to me

<span class="math">\bf{LRN2PROBABILITY~MATH,~N00BZ}[/spoiler]

>> No.3088322

well there are 4 options
BB
GG
BB
GG
BG
GB

it cant be BB

so we end up with
GG
GG
BG
GB

so the chance is 2/4

u mad because you know Im right and people who claim the answer is 1/3 are actually wrong?

inb4 why put 2 GG and BB

because it is 2 different options which both have to be considered G1G2/=G2G1

>> No.3088331

>>3088321
>order is insignificant

I wish more people understood this. I have never seen this question include the order of birth -- that's not to say one couldn't write it as such.

>> No.3088337

>>3088282
the question doesnt say that a SPECIFIC child is a girl, only that between the both of them, at least one is a girl. there are three possibilities that meet the question's criteria:
the elder is a girl, the younger is a boy.
the elder is a boy, the younger is a girl.
both are girls.

those 3 things all have an equal chance of happening, so the odds of both being girls is 1/3

>> No.3088343

>>3088331
this is not a combination problem though, but a permutation one. too bad you fail at logic

>> No.3088349

trolls trolling trolls trolling samefags

its 1/3 if you know anything about probably

>> No.3088353
File: 106 KB, 554x439, 1305528278850.jpg [View same] [iqdb] [saucenao] [google]
3088353

>>3088343
find something better to do than to be horribly incorrect in an obvious troll-attempting manner

>> No.3088357

100% troll, every time.

>> No.3088361

>>3088337
moron detected

by your conventions of language you assign BG/GB as a set of 2, while GG/GG, as a set of one, if you rephrase your english like so

>the elder is a girl, the younger is a boy
>the elder is a boy, the younger is a girl
>the elder is a girl, the younger is a girl
>the younger is a girl, the elder is a girl

you have 50%

>> No.3088362
File: 2 KB, 100x100, fuckingtrollscutitout.png [View same] [iqdb] [saucenao] [google]
3088362

>> No.3088372

>>3088337

wrong.

this has nothing to do with age.

The question asked whats the probability of both being girls. Its 50% It's completely irrelevant whos older and whos younger.

you also forgot

elder girl younger girl
younger girl elder girl.


50%

>> No.3088378

>>3088361
the 3rd and 4th lines are the same thing, thats why they were given the elder/younger specifications. if you wanna be a retard and place PHRASES in different order too, then it would look like this cause you forgot two:

>the elder is a girl, the younger is a boy
>the younger is a boy, the elder is a girl(same)

>the elder is a boy, the younger is a girl
>the younger is a girl, the elder is a boy (same)

>the elder is a girl, the younger is a girl
>the younger is a girl, the elder is a girl(same)
idiot

>> No.3088385

>>3088282
I should elaborate for people who still don't see where "1 in 3" is coming from (or the SUBTLE reason it is wrong for a non-OP person to say).

In the set of parents with two children, approximately 50% have children of the same sex, and 50% have children of different sexes. So in the set of parents with two children 75% have at least one girl and 25% have two girls, and since this two-girl 25% is a subset of the at-least-one-girl 75%, this means that one third of parents with at least one girl have got two girls.

So if you surveyed people on how many children they had and of what sexes, and threw out all of the ones with more or less than two children, and then threw out all of the ones with two boys, and then shuffled what you had left and drew one at random, it would be twice as likely that the randomly drawn sample would show one boy and one girl (2/3rds of the remaining papers) as it would be that it would show two girls (1/3rd of the remaining papers).

However, it was not specified that John Proctor was selected BECAUSE he has two children and at least one a girl. Therefore we default to the case of knowing only that there is one child of completely unspecified sex, and the 50% probability.

>> No.3088401

>>3088385
>one third of parents with at least one girl have got two girls
one fourth of parents with at least two CHILDREN have got two girls. theres the difference. we're performing two stages of "filtering" before looking at the odds:

>having two kids
>not having two boys
the odds of having two girls when the above two conditions have been met is 1/3

>> No.3088402

since were 100% sure that 1 child is a girl the 2nd child only has 2 possibilities:

boy
girl

...i dont see how anyone can really think its 1/3

>> No.3088420
File: 150 KB, 500x500, 1305416746461.png [View same] [iqdb] [saucenao] [google]
3088420

<span class="math">\bf{STOP~BEING~MATHTARDS}[/spoiler]
You imbeciles haven't looked over the very obvious probability that the other child could be a hermaphrodite, which will be represented by H
so the odds are:
{GB, GG, GH} or 1 out of 3.
<span class="math">\bf{OWNED.}[/spoiler]

>> No.3088428
File: 23 KB, 640x480, 1305772684108.jpg [View same] [iqdb] [saucenao] [google]
3088428

>ITT: faggots try to use math to justify their fantasy of switching genders with their sister, calling it the same thing
>nobody fooled except newfags
>mfw

>> No.3088472

>>3088428
You're gay.

>> No.3088482

OP apparently met a man named "John Proctor" and learned about his children, then decided to pose us this problem.

If OP met John Proctor, and it turned out that he had two boys, there would be a 0% chance that he would pose the problem this way, and a 100% chance that he would be asking the question "What is the probability of both children being boys?" instead.

So if:
2 boys: OP will certainly ask "Are both boys?"
1 boy, 1 girl: OP has a 50:50 chance of asking either question
1 girl, 1 boy: OP has a 50:50 chance of asking either question
2 girls: OP will certainly ask "Are both girls?"

Although in 2 of 3 possible cases where OP might decide to ask "What is the probability of both children being girls?", only one is a girl, the question "What is the probability of both children being girls?" is only half as likely to occur in those cases.

Taking into account all available information, the probability is 50%.

>> No.3088531

P(A&B) = P(B)P(A|B)

Conditional Probability

>> No.3088659
File: 11 KB, 251x251, ghuurh.jpg [View same] [iqdb] [saucenao] [google]
3088659

>>3088531
<span class="math">P(A&B) = P(B)P(A|B)[/spoiler]
<span class="math">\bf{CONDITIONAL~PROBABILITY}[/spoiler]
A=probability that the other kid is a girl
B=probability that John Proctor didn't lie about the other kid being a girl
<span class="math">P(A|B)=\frac{P(A&B)}{P(B)}[/spoiler]
<span class="math">P(A|B)=\frac{\frac{1}{2}}{100%}[/spoiler]
<span class="math">P(A|B)=\frac{\frac{1}{2}}{1}[/spoiler]
<span class="math">P(A|B)=\frac{1}{2}[/spoiler]
<span class="math">\bf{PROBABILITY~IS~50\%}[/spoiler]

>> No.3088952

>>3088482
mfw I actually know a John Proctor in real life
And John Proctor is the protaganist in the play "The Crucible"

>> No.3089007

It's 1/2 clearly, ya n00bz

>> No.3089031

> 1/3
> assuming order matters

there are two options
2 girls or
1 girl 1 boy

1/2

>> No.3089622

Anyone who answered 1/2 didn't realize that this was not a conditional probability. You must be careful to read the question.

There is no mention as to whether or not the girl was first, which implies that you shouldn't assume that.

When this troll question is usually posed, it is worded in such a way as to imply a conditional probability.

>> No.3089739

1/3

We know the only possibilities are bg, gb, and gg, bb is not allowed. Since these are equally likely, 1/3.

>> No.3089762

>>3089739
order doesn't matter foo.
1/2.

>> No.3090355

its 50%

>> No.3090383

olookitsthisthreadagain.jpg

seriously cut it out with this shit you don't look smart bragging to underage faggots.