Yesterday there was a thread asking whether Yas Forums was good at math. No one could agree

>first thought is inflicting damage to the butt
200% it is

This is not the same question as yesterday's thread

50% since you only have Group B and Group C to consider?

It’s two thirds chance

66%

75%

100%
Group A never commits crime

The based police force of Randodia uses common sense and goes after the likely criminals so the answer is 95%

Can you explain why it's not the same?

Collaring a criminal is the same as pulling out a green ball

2/3, but clever slide thread.

This is more confusing kek

But there are 2 Bs for every criminal C.

Based and Stopandfriskpilled

Depends if you limp in the group b section of group C into the b group itself, or if you consider them group C still

but group C commits half the crime

no ameritard it's not 50/50, group A never commits crime so they can be omitted from the calculation

the culprit can only be from b or c

Do you want per captia or just a ratio?

They're listed separately aren't they?

3/4

>but group C commits half the crime
No, half of group C commits crime.

CRIME IS FOR NIGGERS

50% unless nigger like OP, then its 100%

I didn't understand it before but I understand it now

Attached: 47b.jpg (665x574, 24.8K)

Okay so obviously group a = white people, group b = niggers, and group c = mexicans.

Think of it like this: there are 6 people: two whites, two niggers, and two mexicans. If someone commited a crime, it was either one of the niggers or the mexican criminal.

The probability would then be (incidences) / (possibilities)

1 crime / (2 niggers + 1 mexican)
= 1/3

Lock up the nigger anyway

Oh good point I should make the skin colors different so it's clear that there's no overlap between B and C

oh wait problem says chance of being in group nigger, so that'd be 2/3

Also this

The crime committing population is 2/3 B and 1/3 C. So the odds the person is in B (assuming we're not including things like odds of false arrests in which case the problem is unsolvable) is 2/3.

There can be false reports filed, police being called to a crime does not mean a crime was committed

Given this, the answer is 33%

two thirds of criminals are in group B, and one third are in group C. You know that the person is a criminal, so the odds that they're in group B are two thirds.

ok here it is with the skin colors changed. hopefully clearer

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66.66666%, because you know it will never be anyone from group A so right off the bat you are at a 50 50 shot from it being either B or C. However you know that of group C, only 50% of them will commit a crime. So you can split them into 2 groups, C.I (Innocent) , and C.G (Guilty)

So now your entire possible choices are going to either be B, or C.G. B is twice the amount of C.G, there for you have a 66.66666% chance of getting group B.

This assumes though, that all groups contain equal number of people, in which case you always know that despite them making up 13% they still commit 50% of the crime.

the answer is niggers

False arrests are not independent from crime committing. Someone who actually commits a crime is more likely to be arrested than someone who doesn't even with the odds of false arrests included.

the answer to this one however, is one-half. There's a very subtle difference. Good bait thread, though.

I was thinking A for Asians but yeah

>implying police never make mistakes
in a perfect world where nobody is ever falsely arrested? 2/3.
in the real world? 1/3.

in either world group b is niggers lol

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it's the same question.

suppose that the groups each have two members - you have one criminal in lockup. what are the chances that the other member of that criminal's race is also a criminal - I.E what are the chances that the criminal is from group B?

can you explain to me what that subtle difference is?

2/3rd

No, it’s 1/2 or 50% chance it’ll be green because since one ball is green, the box with only reds is not possible. So then the two boxes with at least one green ball are the only boxes able to be counted

Only right answer here.

it has to do with the selection. In the box problem, it's worded so that you select a box first. A better metaphor would be this:
Pick a group of people at random (from A, B, or C). Pick a random person from that population, and notice that they are a criminal. What are the odds that every person in that population is a criminal? The answer is one-half because you must have either picked population B or population C at the start.

The original OP wording could be interpreted as this, in which case there's a 100% chance the criminal's in group B. While the answer to the better-worded version is 2/3, there's room for a great math troll in here somewhere.

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