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/sci/ - Science & Math

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3209267 No.3209267 [Reply] [Original]

You have three bunnies, one of which is purple. What's the probability that the fifth one has died less than -20 times?

Bear in mind that these bunnies are not bunnies, but bears. Non-monotonic logic is not allowed to solve this problem.

>> No.3209283
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3209283

100%.

>> No.3209292

0%
Cause of -20

>> No.3209294

>>3209292
That only proves it is 0 or less, not that it is exactly 0.
You can never into quantum mechanics if you don't understand that.

>> No.3209300

>>3209294
You, you dont get it.
-20 times is 20 times living.

>> No.3209303

>>3209267

it depends are these african (Poelagus marjorita) or european (Oryctolagus cuniculus) bunnies

>> No.3209305

50%?
We dont really know, since they aren't even bunnies

>> No.3209307

I have a banana, so your bum is congruent. HA! SIX HUNDRED!

>> No.3209310

>>3209307
Korreckt answer, you divergent alligator!

>> No.3209320

>>3209303
I'd rather assume OP is talking about dust bunnies, or (as implied by "Bear in mind that these bunnies are not bunnies, but bears.") even water bears.
Water bears tend to be not dead 20 times, no matter what you do to them.

>> No.3209333

The less than -20 deaths probability:
If we take -20 deaths = 20 survivals
The probability of it having less than 20 survivals is either 0% or 50% if we assume it speaks for a bunny, since we have 3 bunnies (dead or alive) it means the fifth bunny can be in either state, so the probability of it having less than 20 survivals is like asking what is the chance of it being alive or dead, so we'll assume its 50%.

The purple, the "fifth" bunny and bears are elements of distraction i guess.

>> No.3209347

OP give the answer.

>> No.3209350

It's exactly 0% if we assume that every bunny is granted exactly one life. Proof:
number of times died < -20 ==> number of times lived > 20 which is impossible by assumption ergo the probability is 0. If instead multiple lives are granted then it might change, but we don't have enough information for that.

>> No.3209369

Let's try the numerical approach:
First, iterate over bunnies n times, where n is the number of bunny we are currently "at".
For each bunny, add his name to her number of bears (track the number of bunnies that are bears separately for each color). Then add this value to the number of bears.
This operation was to calculate the value we will use to stop the actual algorithm (I will use pseudocode for better readability):
cur=numberOfElements({<span class="math">x\in\aleph_{sqrt{-1}} : x^{cur} \not! \epsilon[/spoiler]}
while (numBears/bunny[cur].color > <span class="math">\infty[/spoiler])
{
if(bunny[cur].color == (blue && red)) //bunny is purple
bunny[last].died += <span class="math">lim_{0->\frac{\infty}{\not\infty}} \frac{\sqrt{bunny}}{0i}[/spoiler]
}
Then just take the value of epsilon-th bunny. It will be the probability measured in Newtons (you will have to convert to furlongs yourself - I'm not doing trivial stuff for you).

>> No.3209490

>>3209369
and now explain that to a layman.

>> No.3209862

Thats kind of a mindfuck.

I suppose its 33%

>> No.3209876

Three bunnies.

Fifth one.

???????????????????

>> No.3209983

>>3209490
Just rewrite this in INTERCAL, compile against libMonoLogic.so, link with Python interpreter written in Python, add as a binary kernel module loaded before booting, set up true random number generator, run quantum system entangling script, turn off decoherence and run your code.
It's far simpler than it sounds. You would get laughed at for asking this on /g/.
>>3209876
You forgot the essential part:
>Bear in mind that these bunnies are not bunnies, but bears.
Read more carefully next time.

>> No.3210109

>>3209983
you troll?

>> No.3210122

50%, it does or it doesn't. Samey Samey, "scientists".

>> No.3210154

>>3210109
maybe that's an imaginary bunny? I'm either not sure if the "fifth one" was addressing the purple or just any other bunny that is bear.

>> No.3210174

42.