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/sci/ - Science & Math

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9653403 No.9653403 [Reply] [Original]

Well?

>> No.9653405

>>9653403
None of the above

>> No.9653406

What, you want me to fuck you in the ass or something?

>> No.9653407

50%

>> No.9653408

>>9653403
It could be 50% or 0%.

>> No.9653409

>>9653403
2/3

>> No.9653477

Answer is unironically 0%. If it lands on 0% it's still wrong, thus 0%.

>> No.9653481

>>9653477
Being wrong doesn't imply a 0% chance. The chance of it landing on 0% is 25%.

>> No.9653487

>>9653481
Doesn't matter if the chance of it landing on 0% is 25%, the question is how likely is it to land on a CORRECT answer, and none of them are correct, so it's 0%.

>> No.9653500

>>9653403
Fuck you

>> No.9653561

>>9653403
>what is the chance
Lrn2probability fgt pls

>> No.9654311

>>9653487
So you're saying 0% is the correct answer?

>> No.9654317

>>9653403
I usually guess b on tests
Therefore the answer is C because that's wrong

>> No.9654739
File: 656 KB, 3340x3176, the synthesis.png [View same] [iqdb] [saucenao] [google]
9654739

>>9653403
We know exactly one of the answers A, B, C, D is correct as for a multiple choice question. If we assume A is correct, then D is also correct, which means picking a ranndom answer has 50% chance of yielding a correct one. But as both A and D state 25%, this leads to a contradiction. Therefore, neither A nor D can be correct. If we now assume B is correct, then that also leads to a contradiction since the chance of picking B by chance is not 50%. Neither is it 0%, in case of C. And so there is no correct answer.

>> No.9655025

>>9653403
Let's see.
0% isn't correct so that's clearly wrong.
75% chance of choosing the correct answer then.
But you have two 25% which mean you have a 50% chance of choosing those.
Which would make 50% also correct but then the, 25% would be wrong.
When looking at the answers the correct answer would actually be 33%
If you didn't look at the answers and just christmas treed one in it would be 25%

>> No.9655409

>>9653403
A is event of choosing A
same with B, C, D

Y is event of being correct

P(Y) = P(Y|AD)*P(AD) + P(Y|B)*P(B) + P(Y|C)*P(C)

P(Y) = 0*.5 + 0*0.25 + 0*0.25 = 0

>> No.9655434

Since the answer to the question changes based on the answer provided, this allows for a contradiction and therefore cannot be answered, at least where you get a "correct" answer including the 0% answer.

>> No.9655512

>>9654311
Nope, zero's the wrong answer. That's why the odds of picking the right answer are 0%.

>> No.9655550

>>9655512
If the odds of picking the right answer are 0% then C is a correct answer hence the contradiction. Whole problem is fucked

>> No.9655578
File: 1.41 MB, 1433x829, ysbatst.png [View same] [iqdb] [saucenao] [google]
9655578

You just picked the first result, what a lazy sob.

>> No.9655585

>>9653403
There is no such thing as random. Now fuck off.

>> No.9655702

>>9655578
Can you post it in higher resolution, senpai?

>> No.9655723

>>9655702
Google "you should be able to solve this"

>> No.9656476

There are two correct answers, and two incorrect answers, so the answer is B. A and D are correct answer, and B is the probability of guessing the correct answer.

>> No.9657026

>>9653403
Depends on how correct is defined. If there is only one correct letter, then 25%. If the correct is defined by the content of the answer, then 50%.