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Who can do it first?

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youtube.com/watch?v=9mAbsKEyKts
en.wikipedia.org/wiki/Conditional_probability
en.wikipedia.org/wiki/Bertrand's_box_paradox
en.wikipedia.org/wiki/Monty_Hall_problem
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12 you fucking chimp, now go back to /sci/ with your numbers shit

it's actually 24, fuck I'm so tired

y' = 1/2 x^3
1/2 * 4^3 = 32

>(u)

fuck me, i fell for it
fuck you op, saged.

The answer is obviously pineapple.

Is using Reimann Sum overdoing it?

24. They're looking at the derivative of the area under the curve. So that function is its own derivative. Niggerchimps

A 24

I think it's a trick question using two sets of independent/dependent variables
I'm sure you just have to use integration by the disc method using bounds of x=4 and 3/2x^2

It's literally just a "can you read" problem.
A(u) = integral of 3/2 x^2
A'(u) = 3/2 x^2

It's 32

You're an embarassment

Area under a curve is found by integration...

this

24

area under curve is A(u)
derivative of area under curve is the curve that bounds that area, which in this case is y=3/2x^2
sub in 4 for u.

Do your own fucking homework you stupid cunt.

>the integral of a function is the original function
nigga, this ain't e^x

B 32

use the power rule of integration, get .5*x^3, .5*4^3 = 32

it's 24 retards

this is literally kiddy shit, as in chinese kids do this at age 4, please stop embarrassing my great country

Are you retarded or just pretending?
The derivative of an integral is just the original function.

24.

It isn't asking for the derivative of y, it's asking for the derivative of A. A is the integral of y, thus the derivative of y is A. This means it is asking you to find the value of y at 4, which is 24.

Ready for me to post the solution?

>Literal Retards living amongst us
Kill me now

>u=4
>4u

BANE?

I think I might have been to that store.

they are not asking for the area

they are asking for the derivative of the said area, which is the function in question, evaluated at u = 4

24
Back to you go

24... It tells you that A(u) is the integral of (3/2)x^2 so A'(u) is (3/2)x^2.

Dude, you too? I recognize those apples.

WINNER.

24

I came here to shitpost not do math fuck you man

The answer is potato

Me too I other stores they have apples stacked in a different way. Also the fridge with vegetables in the background looks familiar.

It's a big number.

what is this, calculus?

boring

...

it's literally just a matter of plugging 4 into x.

How do you retards get this wrong?

...

ITT: Sup Forums learns the fundamental theorem of calculus

>tfw cellular metabolism test in the early morning
>tfw i'm helping user with his homework instead of sleeping

2/3

1/3 1/3 1/3 1/3 1/3 1/3

So if the boxes are A, B, and C
If you grabbed a gold ball at random, there is a 2/3 chance it was from A, 1/3 from B, and 0 from C.
So of that 2/3, there is a 100% chance the next will be gold, and from 1/3, a 0% chance.
So 2/3.

2/5

You won't need sleep if you've memorized all your cycles.

You did memorize your cycles, right? Oh hell who am I kidding you can just throw ATP in there somewhere and probably walk away with half credit.

1/2

It's 1/2. You pick a ball from the box and it's gold. That means you pulled from either the middle or the left box. If it's the left box, the next ball will be gold; if it's the middle, it's silver. Ergo, 1/2.

No, no. it's 2/3.

ITT: Dude gets people to do his homework for him.

t. someone who failed probability

Look at it like this. Of the 6 balls, you have an equal 1/6 chance of picking any of them.
So a 2/6 chance of picking a gold ball in the first box, a 1/6 chance of picking from the second box, and the rest we can discard because it doesn't fit the condition that we got a gold ball.

You've picked a gold ball removing it from the equation. You can't look inside the boxes to see which one you have pulled from. Five balls are left, two gold, three silver.

oh great i walked into another faggot thread
youtube.com/watch?v=9mAbsKEyKts

50% because you already eliminated the third box with your first pick

50%

you wouldn't be taking out of the box with 2 silvers

it's 50%

Ivan is right, guys.

It's 24 and you can do it in your head when you realize that the derivative of an area like this is just the upper bound graph that defines it. So it's literally plug 4 into the graph formula.

Please stop making yourself look like an ass even if it's just pretending.
I have a math degree and can safely say that it's 2/3.
You can solve it either via exhausting+eliminating combinations that fail the "you picked a gold ball" requirement, or by doing it the easy way as in

24

It follows directly from the fundamental theorem of calculus, which says that if f(x) is a line, and A(x) is the area under that line above the x-axis, A'(x) = f(x).

the question doens't ask about the first pick. It asks about the NEXT pick from the same box. There is no 3 at all in the equation. The NEXT pick is either silver (box C), or gold (box A)

>American education
Lol

We're looking for the conditional probability of having chosen a ball from the first box with the condition that the ball we've chosen is gold.

P(A\cap B)=1/2
P(B)=1/3
P(A|B)=(1/3)/(1/2)

This can, of course, be easily seen directly but that's a simple way of verifying the answer.

en.wikipedia.org/wiki/Conditional_probability

Lurk it.

It's 1/2 my friend. The question tests if you can recognize whether these events are independent.

You have one golden ball in your hand. There are only 2 boxes with any golden balls. If you have box 1, you'll get a golden ball. If you have box 2, you'll get a silver ball. The probability is 1/2 or 50%

No we do not, we already picked a gold ball.
50%

Yeah this makes the most sense. It's kind of like the 2nd tier of a probability tree. We already know the past.

1/2 it is.

I'll do this the long way since you clearly don't understand.
Call the boxes A, B, and C, and the balls in each box A1, A2, B1, B2, etc
So A1, A2, B1 are gold, the rest are silver

So if you grab two balls from the same box, there are six ways to do it, all with equal probability assuming the box was chosen randomly:
A1 first, then A2
A2 > A1
B1 > B2
B2 > B1
C1 > C2
C2 > C1
Of course, the latter three can be eliminated, because those do not involve picking a gold ball first.
That means we're left with:
A1 > A2
A2 > A1
B1 > B2
The first two result in a gold ball being picked second, the third does not. All three still have equal probability.

Reading for you math failures: en.wikipedia.org/wiki/Conditional_probability

They aren't independent.
It's asking about the chances of the second ball being gold GIVEN that the first one was gold as well.

Look at it this way: what if each box had a hundred balls. The first box is 100 gold, second is 1 gold, 99 silver, and the third is 100 silver.
If you grabbed a gold ball on the first grab, there would only be a 1/101 chance it came from the second box.

The condition is that the first pick was a golden ball picked from one of the first two boxes. Therefore the next pick has an equal chance of being gray or gold. This is presuming one knows the facts of the first paragraph.
>Russian education

You could've saved yourself some headache and just linked the paradox's page directly.

en.wikipedia.org/wiki/Bertrand's_box_paradox

What's the odds that you were wrong on your first pick??

The odds that you were wrong on the first pick is the same as the odds that it's in one of the doors you didn't pick.

The derivative of the integral of a function is just equal to the function, regardless of the original function.

I[D[f[x],x],x] = D[I[f[x],x],x] = f[x]

So if A is the integral of y, then A' is y

So A'[4] = y[4] = (3/2)(16) = 24

>all these Americans are retards memes I see on Sup Forums being pushed every single fucking day by yuroshits
>American tricks yuroshits into doing his homework for him
>who's the retard now
>mfw

You're doing the math as if we hadn't already picked a golden ball. We have one of them in our hand. Thus the second ball is either golden or silver, or 50/50 odds

Last attempt. Yes, we KNOW that the ball we've picked is gold and, with that taken into account, we're interested in the probability of this ball being from the first box. This is, by definition, the conditional probability I've mentioned.

(I really hate using this as an argument but I am a mathematician, i.e. I do pure mathematical research for a living.)

That's a pretty easy integral to do mate, I just woke up, 3/2int(x^2) = (3x)^3/4 now just plug in x using the fundamental theorem of calculus and you have your answer

what is the purpose of calculus in the real world?

Oh wow. You're right. I didn't want to think this was just like that goat/car question behind three doors, but it is.

So basically, knowing that we picked out a gold ball, chances are we're grabbing from the 2-gold box, increasing the chances of drawing another gold.

a u

high level problem solving. not to be a great mathematician, but to be able to come up with strategies for complex problems.

Tfw when Sup Forums did a Calculus problem while /sci/ would just call you a dumb nigger and not solve it.

like what?

1/2

Yes, but knowing that we picked out a gold ball, chances are it was from the 2-gold box.

And that means we're more likely to draw another gold.

...

>we're interested in the probability of this ball being from the first box

It's 50% because there is no golden ball in box C.
You either picked box A or B.

>I am a mathematician

but what if every mathematician is wrong on that one?

idk tension simulations on a bridge? Aerodynamics, computer science, flight models

I don't think anyone in a industrial field is gonna give a doodoo if you memorize euler's number or some nonsense, but you need to be able to put formulas and functions together (or make your own) to solve problems.

>all these people saying 50%

fuck I just got cucked by math, this is just a monty python problem in disguise

24

Now take your retarded americlap math and shove it up your ass

>monty python problem
it's called the full monty problem you reatard

michael hall problem strikes again.

wtf i hate that guy now

>but what if every mathematician is wrong on that one?

>racist white patriarchal science

why do your own work when you can trick (use social skills) someone with better logic skills into doing it for you

It's called the Monty Hall problem is it not?

It's called the Monty Hall problem You faggot

en.wikipedia.org/wiki/Monty_Hall_problem

its 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

What if y'all are against integration?

You know you can test this yourself. Record the frequency and calculate for a sample probability.

you aren't replacing the ball. It's 50/50