Biz can never solve this, and this is why biz will never make it

*1. 66%

The possibilities are:

GG
BG
GB

Correct? BB is ruled out in both scenarios because the daughter answered the door? And since GB and BG are the same, that leaves GG and BG(GB) which is 50%

>t. brainlet

No gender is ever mentioned so it's 50% both answers.

FPBP OP BTFO

That's basically what I said

(i) 50%
(ii) 25%

You're getting memes because your bait is so bad you can't even bait the lowest IQ board on 4channel.
Reasses your life choices.

Not just that but we don't know if the daughter identifies as a daughter, son, aunt, uncle

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

FPBP OP BTFO

(i) 1/2
(ii) Intuitively I feel it's 1/2, there could be some monty hall element to this making it 1/3 but I don't think so.

FPBP OP BTFO

FPBP OR BTFO

FPBP OP BTFO

Why do these retarded ass mutual exclusion problems always surface such brainletry on Jow Forums.

>goes on biz
>doesnt ask a biznes or finance question
>upset that no one answers

FPBP OP BTFO

The gender of one child has no influence on the gender of the other child. It's 50% in both cases. This one was boring.

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FPBP OP BTFO

1 is E(x) or ~63%

FPBP OP BTFO

thanks just bought 100k

This is correct

there are more than 2 genders, so its endless.

FPBP OP BTFO

FPBP OP BTFO

What kind of reddit tier replying bullshit is this

Jow Forums is 90% reddit refugess. Didn't you know that?

FPBP OP BTFO

I) 50%, other child could be a boy or a girl
II) 50% again, other child could still be a boy or girl

Is biz really that retarded?

I think there is a monty hall thing going on with the second part. First is clearly 50%.
G -> G
G -> B
B -> G
B -> B
Lower two are eliminated since we know the oldest is a girl.
G -> G
G -> B
Since we don't know if the daughter answering the door is younger or older the above possibility has a 100% rate of showing girl while the below has a 50% rate of showing girl. Like the balls puzzle this means 2/3 chance the other daughter is a girl or 50% for part 1 of the question, 66.7% for part 2.

Of course this is disregarding the 90quintillion possible genders.

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Correction the question is asking for the chance that the other child is a boy which means 1/3 for the second part

You know for a fact they have a girl so if it's the younger daughter who opens the door then it's 100% the other child is also a girl.

In the first one it doesn't matter the gender of who opened the door.

Jesus dude I really hope you're trolling

Nah math is retarded. the modern accepted answer is something else.

We don't know the oldest is a girl

0%, nobody who was married with children would ever invite me over to their house

FPBP NIGGERS BTFO

i and ii and independent.
The fact that the daughter is eldest is irrelevant information. The fact that a daughter opens the door is irrelevant info. The fact that there's 2 kids is irrelevant info. Is a child born boy or girl is an independent event unrelated to anything else.
There's so much irrelevant info here it's clouding your judgment.

The question is basically "what is the probability a child is a boy". The question is basically the same for both i and ii. The child can be boy or girl.

50%

the absolute STATE of Jow Forums

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Look up Monte Hall, gets people completely convinced the answer is obvious this is a rehashing of the problem with 3 pots and 2 gold balls in 1, 2 silver balls in 1 and a gold and a silver in the third. You think the answer is obvious so you underthink it.

Think of it this way, if the house has 10 million boys in it and one girl or 10 million girls in it and no boys and you open the door and see a girl what are the odds you're in the first possibility as opposed to the second?

What are you talking about lul

Probably

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There are 4 ways you can have two kids:
1. boy, then another boy.
2. boy, then girl
3. girl, then boy.
4. girl, then another girl.

i) A girl answered the door, so case 1 is eliminated. In what is left, you have a 2/3 chance of a boy.

ii) The eldest is a girl. That means only 3 and 4 applies to the situation. In which case, the other child has a 1/2 chance of being a boy.

And that's my final answer.

i) 2/3

ii) 1/2

I'm glad there are so many dumbfucks on /biz. If you guys weren't here, I wouldnt be able to make my money

You're making the problem more complicated than it is. Since you don't know whether the girl who answered the door is the older your younger sibling, you have zero information pertaining to the gender of the other sibling. Therefore it must be 50%.

Explain then, since all you can do is answer with jokes and memes

This is what tricks people they think that age can't be relevant information. Yes the sex of the child is an independent event, but we don't know the sex of the child, and the likelihood that a person is one thing or another changes with circumstantial evidence. If the oldest child is a girl then the youngest must either be a boy or girl with equal likelihood.

GG vs GB

However while the likelihood of either scenario occuring is equal, there is a 100% chance that a girl will answer the door in scenario A and only a 50% chance that the girl answers the door in scenario B. Since a girl has now answered the door it is slightly more likely that we are in scenario A.

Unfortunately this is too difficult for low-normal IQ types to comprehend so every time one of these threads gets dropped you just have to accept that brainlets aren't going to be able to understand probability.

Thank you for this explanation.

This is really amusing

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I know you think that, but if you just look this up you'll see many very good explanations. You think i'm making it more complicated than it is because you're not using the available information to derive likelihood.
>You have zero information pertaining to the gender of the other sibling
We don't know which sibling we're looking at to begin with and you don't need to KNOW which sex the other sibling has, the question isn't which sex IS the other sibling it's asking how likely they are to be one sex.
>Therefore it must be 50%
"We don't know therefore it must be 50%" is like saying that you have a 50% chance of winning the lottery because you don't know if you'll win or not.

I simply don't understand how the sex of one child affects the potential sex of the other.
The probability of a child being a boy is 50%. The fact that they have another child who may be a boy or a girl is irrelevant. The fact that we know they have a girl in scenario ii is irrelevant. It's not like after that had a daughter, they were more inclined to have a boy next time around.
Imo the sex of a child is completely independent of everything which is why I think the answer for both scenarios is 50%.

But it's more likely that you're looking at a 2 girl house because a girl answered the door. If the door wasn't opened then it would be 50/50 but if we've seen a girl then there are 3 possibilities.
1. We're seeing the older girl in GG
2. We're seeing the younger girl in GG
3. We're seeing the older girl in GB
And all of these possible situations are equally likely.

>We don't know which sibling we're looking at to begin with and you don't need to KNOW which sex the other sibling has, the question isn't which sex IS the other sibling it's asking how likely they are to be one sex.
They are mutually exclusive.

>"We don't know therefore it must be 50%" is like saying that you have a 50% chance of winning the lottery because you don't know if you'll win or not.
If the lottery was a coinflip then yes, you would have a 50% chance of winning. Gender is a coinflip, and one child's gender does not influence the other child's gender.

Think of it this way, you have two households.
H1 contains 1million girls and 0 boys
H2 contains 1million boys and 1 girl
You open a door and a girl answers
Are you as likely to be in H2 as H1?

Seriously look up Monte Hall, you still probably won't be able to understand but atleast you'll be worth replying to this shit is why this whole forum are cryptogamblers

I see you're talking about question i).
Why are you counting GG twice? It's only 1 item in the sample space. And you forgot BG.
So it would be GG, GB and BG in the sample space.

The fact that we've seen a girl is irrelevant. The gender of the next child has nothing to do with the gender of the first child. No doctor in a hospital will ever say "what's that, you've had 4 daughters in a row? well then it's time you've had a son this one will definitely be a boy." Sex of a child is a coin flip, 50/50. Everytime.

That's a screwy scenario. I know it's just an example but the odds of having 1 million girls and 0 boys is (1/2)^1000000. This would never happen. There should be about 500k boys and 500k girls, in which case it's about 50/50 what gender opens the door.

That scenario literally has nothing to do with this question. We know there are only two siblings, and we only know that one of them is a girl.

I know what the monty hall problem is. It's unrelated because we only have two options here, not three. Now go look up gambler's fallacy.

This thread is the final blackpill. Jow Forums has been taken over by faggots and crypto is dead. I finally have the courage to leave.

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Both are 50%.

You are all fucking brainlets. No wonder you lost all your money on shitcoins.

There are 4 possibilities
GB, older answers door
BG, younger answers door
GG, older answers door
GG, younger answers the door
2/4 of those possibilities leads to the other being a boy.

>The sex of a child is a coin flip
Yes I am aware and have not once disagreed with this no I have no idea why you keep bringing it up.

It's the perfect scenario for this example, it demonstrates that if there is a higher percentage of girls in one theoretical house you are more likely to be looking at that one. Saying "this would never happen" does not change the validity of these circumstances and while you can't understand the others I know you can understand these.

Saying "this would never happen" is a cop out in a math problem and you know it. I'm just demonstrating how when you change the percentages even further the gap in likelihood gets even bigger.

I think i'm leaving too, there used to be some really bright people on here but these days they can't even look up the answers to math problems and shit on google, the advice here was never really worth trusting but it really is a reddit.biz now.

Both questions are about an unknown with two outcomes. We have no information relevant to the probability of the outcome of the unknown.

It's not a monty hall problem. It's a disguised coin flip problem. 50%.

>Hint: Define your sample space carefully!
Guys ffs i can't belive you didn't see the OBVIOUS clue! 'A married couple has 2 children whom they have not met'
Its doesn't say they are together, if they haven't met there kids it means only 1 thing!

The kids are the children of niggers! Who fuckin cares about niggers!

Your welcome OP, this /thread is now closed!

>one coinflip influences the next coinflip
>this person thinks he has a superior IQ

You're a brainlet See

Omg finally someone with triple digit IQ

Retards.
analyticsvidhya.com/blog/2017/04/40-questions-on-probability-for-all-aspiring-data-scientists/
>2) Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl?

>You can assume that there are an equal number of males and females in the world.

>A) 0.5
>B) 0.25
>C) 0.333
>D) 0.75
>Solution: (C)

>The outcomes for two kids can be {BB, BG, GB, GG}

>Since it is mentioned that one of them is a girl, we can remove the BB option from the sample space. Therefore the sample space has 3 options while only one fits the second condition. Therefore the probability the second child will be a girl too is 1/3.

Retard.
en.wikipedia.org/wiki/Boy_or_Girl_paradox
>However, if the family was first selected and then a random, true statement was made about the sex of one child in that family, whether or not both were considered, the correct way to calculate the conditional probability is not to count all of the cases that include a child with that sex. Instead, one must consider only the probabilities where the statement will be made in each case.[12] So, if ALOB represents the event where the statement is "at least one boy", and ALOG represents the event where the statement is "at least one girl", then this table describes the sample space:

Equal occurence rates for all 4 scenarios because 50/50 on the sexes as brainlets love to say.
GG
GB
BG
BB
#1 50% boy in house
GG: .5 occurence rate of G
GB: .25 occurence rate of G
BG: .25 occurence rate of G
BB: Ruled out
#2 33% boy in house
GG: .67 occurence rate of G
GB: .33 occurence rate of G
BG: Ruled out
BB: Ruled out

Read this example again if you don't understand how you can be more likely to be in one situation as opposed to another based on what you are observing. Look up the boy girl paradox on wikipedia this is a rehashing of that for less ambiguity. An investing forum that doesn't understand probability is not a place worth spending time.

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lul someone who has a clue finally showed up right as I was finishing good luck man

Brainlet.

Brainlet.

>An investing forum that doesn't understand probability is not a place worth spending time.
Sad and true. 2019 will be the year of nobiz for me.

wew, i came to say 56%, but this is way better.

But why are the different combinations of BG relevant?

An unknown human has a 50% chance either way.

You know for a fact the oldest ISNT a boy. Your calculation has to include this information.

FPBP OP Eternally BTFO and Jow Forums redeemed

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>But why are the different combinations of BG relevant?
They're not. This is what happens when people don't have an intuitive understanding of probability. They blindly follow formulas that they don't actually understand.

It's irrelevant.

But all that tells us is that the other child is younger. There's noting to calculate?

FPBP OP BTFO

1) 50%

2) 33%

G-B
G-G

Three different chances for a girl to open the door, with only one of them leading to a boy sibling. Therefore it's 33%.