Biz can never solve this, and this is why biz will never make it
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50/50
50 percent for both you faggot
Not doing your homework faggot and I’m not interested in being an actuary.
Sage
49% and 51%
based af
checked
Assume that the family is hispanic. They have 20 children you have met and the wife wreaks like tortillas fried in lard. What are the chances that the daughters are all sluts?
100%
Extremely easy
First scenario having see one daughter, the options are
FF
FM
MF
therefore probability of a male as other child is 2/3
Second case, we have an ordered pair, so the only choices are
FF
MF
so it's 1/2
Wouldn’t you need census data or something to get anywhere close?
checked
I. 1 in 2 EZPZ
II. 3 possibilities
Can't be BG
GB
Gg
gG
1 in 3
t: brainlet
>he thinks smart people make it
Sorry to burst your bubble broski but we're living in a clown world where only luck and slyness pays off. Well, at least you're not a "hard work" cuck.
Wrong
Wrong
Wrong
Correct
The question presumes that each event in the sample space is equally probably, so you're right in a practical sense, but not in the context of this question.
But what about the beaner sluts?
Oh, the answer was all of them. Presumed it didn't require confirmation.
>Pic related
What are the odds of pregnancy if you bang a beaner slut without a condom?
im not too sure about that one mate but the chance of getting stds is 100%
100%?!?!?! God damn. What if she's got two kids already but still looks good and keeps asking you to come over and do her in the butt? What are the chances you'll get an STD?
I don't understand. Maybe I'm a brainlet?
What's the difference between MF and FM?
either bait or you're as derpy as the brainlets who bet red on roulette after a black shows up
flip a coin twice, there are 4 possible outcomes
HH
HT
TH
TT
Don't they teach basic binomial probability where you are from?
I always just go to the craps table and bet the horn.
Explain your logic and be ridiculed, brainlet.
The chances of getting either red or black are the same, but the chances of getting 2 reds and 2 blacks out of 4 rolls is not the same as getting 4 reds. This is because there are far more result permutations that can lead to 2 reds and 2 blacks. think a bit m'negro
Why cant FM happen in the 2nd case?
Change it up user. You already posted this before
Right, the chance of each set of possible rolls is 1/4. However, in this case, there are two coins, one of which is already flipped, and reveals, say, heads. The second is yet to be rolled, and the possibility that it will be heads or tails remains the exact same: 1/2.
4 possibilities:
FF
FM
MF
MM
However each 'M' or 'F' is independent of its set. The first one appearing as F doesn't effect the chances of the second appearing as either M or F.
Am I missing something here?
where can i see this bitch naked
I don't know. I think Merkel has her locked up in a reeducation camp
What is this nu math horseshit?
If you flip a coin and its heads, this does not somehow change the probability that the next flip will be tails to 66%.
Oh my god. 42 year old boomer here and I have never seen such madness. Is this why millenials and zoomers are retarded?
Actually, it's not even like that. MF and FM are the exact same thing.
Given that in a set, the first one isn't representative of who answers the door, the possibilities are:
FF
FM
MM
That still doesn't change anything. Even if the female answers the door, the probability of the second child being male or female remains exactly the same. The probability of how the set manifests itself isn't the same as the probability of what gender one or the other child is. Each is independent of one another.
>First scenario having see one daughter, the options are
>FF
>FM
>MF
>therefore probability of a male as other child is 2/3
Wrong, because the posterior probability of FF is double that of each the MF/FM. Hence for both questions the answer is 50%.
The first question is sometimes asked ambiguously, but in this case it is not at all ambiguous.
en.wikipedia.org
Stop being a nigger with these memorized answers to trick questions and think them through for once.
How many of their children have I already met, and what are their genders?
You've met the husbands 8 step children. They're all downies with tree trunk necks
2 children, birth possibilities are:
fm
mf
ff
mm
can't be mm, since girl answered the door. we're left with fm, mf or ff. f answers 1st, (f) then m or (f) then m or (f) then f
2/3 chance boy
second part of question
fm (can't have mf since girl oldest)
ff
(can't have mm since girl n
answers door)
1/2 chance boy
Keep in mind that the lady of the house may have answered the door, since she is also a daughter.
>dumb faggots thinking its the monty hall problem
MF
FF
MM
FM
FM is equivalent to MF, as the order of the children is not relevant.
That leaves two possibilities, {MF, FF} 50% assuming each gender is equally probable
>>dumb faggots thinking its the monty hall problem
ITT this
So many people looking like dumbasses in this thread. You know there's a wikipedia page that can explain this to a brainlet, right?
OMFG.
Mind = blown.
Another garden shed thread that I can't profit from. Thanks OP.
>Keep in mind that the lady of the house may have answered the door, since she is also a daughter.
It really triggers me that in 2019 you'd assume that one of the parental units must be female.
there's a 50% chance the firstborn will be a girl and a 50% chance the firstborn will be a boy and a 50% chance the secondborn will be a girl and a 50% chance the secondborn will be a boy
multiply 50% by 50% you have 4 possible permutations with a 25% chance each
mm
mf
ff
fm
the order is relevant dumbass you can't combine mf and fm because the numbers 0.25 =/= 0.5 you can't make those numbers equal
Touche
en.wikipedia.org
surely, Jow Forums isn't this stupid is it?
(ii) is 100%. Think about it, if someone were to tell you that the eldest is female that obv implies the younger is male. Your probabilistic garbage gets btfo'd in the face of common sense. Sage
the problem didn't ask anything about age. learn to read before you call others dumb
>calls people dumb faggots
>don't call others dumb
kys faggot
also this has already been proven to be a meme paradox by the posting of the wiki link so the arguing is just for shits and gigs
nobody is right. we've been bamboozled
13% for (i), 50% for (ii)
>the absolute state of Jow Forums
I didn't even go to college or take probability in high school, but this is easy as fuck.
The first child is a loli, the second is undetermined, thus splitting the timeline into two timelines which are equally likely as of right now
>loli - loli
or
>loli - shota
Next, the door is opened, revealing a loli. This gives us more information as to which of the two timelines we're more likely to be in.
GIVEN we are in the double-loli timeline (LL), the door opening would reveal a loli 100% of the time.
GIVEN we are in the loli-shota timeline (LS), the door opening would reveal a shota 50% of the time
We now know we are twice as likely to be in the LL timeline versus the LS timeline. This means we have to have a 66.66% chance of being in the LL timeline than the LS timeline (33.33%)
And there is our answer. There is a 33.33% chance the second child is a shota.
Don't reply if you didn't get a perfect or near perfect score on your ACT.
I think it's 1/3 chance.
Let A = other child is a boy
B = daughter answers the door
We need to find A given B (A/B)
(A/B) equals (B given A) times A divided by B (conditional probability law)
(B given A) means daughter answers the door given that the other child is a boy. This is a simple 50/50 since if one is a boy and we know eldest is a girl there's an equal chance.
A is 50/50 as established in the first problem
B: One child is a girl, one is 50/50. So probability of B is 0.5*1+0.5*0.5=0.75.
Then probability of (B/A) is 0.5*0.5/0.75 = 1/3 or 33%.
>The question presumes that each event in the sample space is equally probably, so you're right in a practical sense, but not in the context of this question.
No you are just a faggot, the question is simple in both cases you are asking what the probability that a person is male or female. You are overthinking it with all of your faggotry.
>muh sample size
>muh technicalities
Nobody likes you and you know why but you refuse to change.
2nd part
You know it is either FM of FF. 50% chance of either. If it is FF then there is 100% chance a girl opens the door. FM then 50%. So wouldn't it be 25% chance boy?
Your head is in the right place, you just wound up with the wrong answer. 100% is twice as large as 50%, whereas 75% is 3X as large as 25%. The answer you're looking for is 33.33% (half of 66.66%)
I'll give you a pass though, considering how much the average person apparently struggles to even get that far
The mother is also someones daughter
Dude there's a reason why you're poor.
If you literally recreated the problem conditions and went to multiple houses where you knew one child was a daughter and a daughter opened the door, you'd end up with a 33% chance. This is the literal probability. Fuck sample size fuck technicalities. I mean...this is the actual fucking probability, you're just too stupid to wrap your mind around it. This user explained it best:
or if you want pure math I gave a statistical explanation, read this you brainlet
And getcho head straight, you can't amount to much in life if you refuse to learn and admit your mistakes
(i) 4 Outcomes, B/B B/D, D/B, D/D. Given that we know one of the kids is D, that means the 3 outcomes are now B/D, D/B, and D/D so there's a 2/3 chance the other kid is a boy.
(ii), Outcomes are the person who answers the door is the Older Daughter which means younger sibling could be either sex, or person who answers the door is the younger sibling which means it has to be 2 Girls. This means there are 2 outcomes where the remaining sibling is a girl and one outcome where they're a boy, so there's a 1/3 chance the other child is a boy.
End of the day roulette isnt 5050 because of 0 and 00. This is what gives the house its edge.
>1/2 = normal people
>1/3 = faggot commies
If B=boy G=girl, and we assume equal chances (50%) for both, then the sample space in instance i) is:
BB BG GB GG
therefore the chance is 50%.
However in case ii) you know there is at least one G and the space becomes:
BG GB GG
Therefore the chance become 66%
shit I'm wrong, yeah this one is correct
>bg gg
I did say it was an assumption faggot