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> However, I enjoy calculus and have taken advanced college courses on the subject and would be happy to help with any specific questions/problems in the future if need be.
Can you explain fractional derivatives to me? e.g., what it would mean to take something to the 1.5th derivative?
Also, why isn’t “edit post” wedged between “flag post” and “quote post”?



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> *Originally posted by **[LukeMann](/forums/2/topics/73132?page=17#posts8184338):***
> > However, I enjoy calculus and have taken advanced college courses on the subject and would be happy to help with any specific questions/problems in the future if need be.
>
> Can you explain fractional derivatives to me? e.g., what it would mean to take something to the 1.5th derivative?
>
> Also, why isn’t “edit post” wedged between “flag post” and “quote post”?
Fractional derivatives are something I do not have much experience as its applications are not of use in my field. From my limited understanding it is similar to the process of calculating a standard derivative except that the integer is replaced with a variable and calculated by (d^a/dx^a) x^a = k ![ka](/ "ka") x^ka.
The above equation is tough to read on here, but lets say you want to find the derivative of some function to 2.5, then what I believe you can do is find the second derivative as usual and then find the half derivative using the above equation and multiply those two answers together. There are whole courses that can be taken on this subject of which I have not taken, but I hope this may clarify it a small bit at least.
I would suggest researching a bit to make sure what I remembered is correct haha, I believe it is but I’m not 100%.



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> *Originally posted by **[dEsIrEcAt](/forums/2/topics/73132?page=17#posts8185661):***
>
> 0+0=?
>
> The hardest math question.
>
> I can’t even solve it lol.Why, my dear boy. The answer is obviously 0+0= the size of dEsIrEcAt’s penis!
(It amazes me how many scrubs still attempt to troll this thread after 4 years)



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> I’m going straight from memory here but from what I recall pre calculus essentially is a bridge between what you are doing now and calculus.
Man, you know ily Braves.
But no shit.



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If there are three bags of coins. One with two gold ones, one with a gold and a silver coin, and one with two silvers. If you pull a gold coin out of the bag, what is the probability that the next coin will also be gold?



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> *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186026):***
>
> If there are three bags of coins. One with two gold ones, one with a gold and a silver coin, and one with two silvers. If you pull a gold coin out of the bag, what is the probability that the next coin will also be gold?
Okay well based on the wording I’m under the assumption that each bag has 2 coins. So for your first bag you have a 100% probability, the second bag you have a 50% probability, and the last bag it’s a 0% probability because there are no gold coins in that bag. To average these bags (to get the mean) you simply add them all together, so 150%, then you divide it by 3 (how many percentages you’re averaging), and that will come to 50%. So the answer should be 50%



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> *Originally posted by **[GRAPHICDESIGNC](/forums/2/topics/73132?page=17#posts8186032):***
> > *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186026):***
> >
> > If there are three bags of coins. One with two gold ones, one with a gold and a silver coin, and one with two silvers. **If you pull a gold coin out of the bag, what is the probability that the _next coin_ will also be gold?**
>
> Okay well based on the wording I’m under the assumption that each bag has 2 coins. So for your first bag you have a 100% probability, the second bag you have a 50% probability, and the last bag it’s a 0% probability because there are no gold coins in that bag. To average these bags (to get the mean) you simply add them all together, so 150%, then you divide it by 3 (how many percentages you’re averaging), and that will come to 50%. So the answer should be 50%
Word problems are a bitch! Ya think ya know what they’re asking and BAM! Brick wall with spikes and organs.



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> Word problems are a bitch! Ya think ya know what they’re asking and BAM! Brick wall with spikes and organs.
My problem with word problems is that they are fucking retarded and never provide enough information.



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> *Originally posted by **[GRAPHICDESIGNC](/forums/2/topics/73132?page=17#posts8186070):***
> > Word problems are a bitch! Ya think ya know what they’re asking and BAM! Brick wall with spikes and organs.
>
> My problem with word problems is that they are fucking retarded and never provide enough information.
I say when it comes to those guys read em thrice. Half the problem of a word problem is interpreting said problem correctly.
Unless you’re doing SAT stuff, then you read half the question and make and educated guess.



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here guys
G = gold
S = silver
GG
GS
SS
those are the contents of the three bags



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> *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186079):***
>
> here guys
>
> G = gold
> S = silver
>
> GG
>
> GS
>
> SS
>
> those are the contents of the three bags
50%



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> *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186079):***
>
> here guys
>
> G = gold
> S = silver
>
> GG
>
> GS
>
> SS
>
> those are the contents of the three bags
If you are pulling coins from the bags at random, since you have already pulled a single gold coin, that probability of you taking another gold coin is 2 (gold coins left)/5 (coins left overall), or 40%.
However, this problem seems to imply that you picked a single bag to pull both coins from (that actual word problem in your book should specify this better), in which case things are different. From the statement of what is in each bag and what we know of our first draw (that you’ve already pulled a gold coin), the potential contents of each bag is now:
Bag 1: \*G
Bag 2: \*S
Bag 3: SS (cannot have been bag 3 since we know you already got a gold from it)
giving you a 1 in 2 chance, aka 50%.



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> *Originally posted by **[braves055](/forums/2/topics/73132?page=17#posts8186301):***
> > *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186079):***
> >
> > here guys
> >
> > G = gold
> > S = silver
> >
> > GG
> >
> > GS
> >
> > SS
> >
> > those are the contents of the three bags
>
> 50%
It’s 2/3, isn’t it? Let’s relabel these coins for distinction
G\_1, G\_2
G\_u, S
S, S
So there are equal chances that the first coin you drew is G\_1, G\_2, or G\_u.
If the first coin is G\_1, then the other coin is G\_2 which is golden.
If the first coin is G\_2, then the other coin is G\_1 which is golden.
If the first coin is G\_u, then the other coin is silver, which is not golden.
Thus 2/3 circumstances of drawing a gold coin first result in the other coin being golden.



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> *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186026):***
>
> If there are three bags of coins. One with two gold ones, one with a gold and a silver coin, and one with two silvers. If you pull a gold coin out of the bag, what is the probability that the next coin will also be gold?
Define “the bag”. Also if we’re going to pull out of the same bag or not.



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> *Originally posted by **[LukeMann](/forums/2/topics/73132?page=17#posts8186918):***
> > *Originally posted by **[braves055](/forums/2/topics/73132?page=17#posts8186301):***
> > > *Originally posted by **[FULLMEASUREZAM](/forums/2/topics/73132?page=17#posts8186079):***
> > >
> > > here guys
> > >
> > > G = gold
> > > S = silver
> > >
> > > GG
> > >
> > > GS
> > >
> > > SS
> > >
> > > those are the contents of the three bags
> >
> > 50%
>
> It’s 2/3, isn’t it? Let’s relabel these coins for distinction
>
> G\_1, G\_2
>
> G\_u, S
>
> S, S
>
> So there are equal chances that the first coin you drew is G\_1, G\_2, or G\_u.
>
> If the first coin is G\_1, then the other coin is G\_2 which is golden.
>
> If the first coin is G\_2, then the other coin is G\_1 which is golden.
>
> If the first coin is G\_u, then the other coin is silver, which is not golden.
>
> Thus 2/3 circumstances of drawing a gold coin first result in the other coin being golden.
Your draw pool is incorrect as you did not factor in the bags correctly, look at what i did above you.
You have already drawn a gold coin.
This eliminates the S, S bag from contention leaving us with 2 bags to draw from.
The first bag has another gold coin, the second bag has a silver coin. As you are drawing from only 1 bag, you have a 50/50 chance of it being the bag you need.
To explain what you did, you calculated the odds of drawing another gold coin from a bag that contains another 2 golds and one silver.



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he’s still not wrong
assuming you picked one bag randomly and picked up a gold coin from it:
there are 6 possible outcomes
choose bag 1: G1/G2 or G2/G1
choose bag 2: G/S or S/G
choose bag 3: S1/S2 or S2/S1
you either picked the first gold coin in the first bag, the second gold coin in the first bag or the gold coin first in the gold/silver bag first
since we didn’t pick a silver coin first, we can eliminate 3 of the 6 possible outcomes
3 possibilities left, 2 of them will give you a second gold coin



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As I said, that’s not correct for this problem, you are finding the likelihood of drawing a gold coin at all, but what the problem is asking is not the odds of drawing a gold coin, but rather that the NEXT coin will be gold which is dependent on us having bag 1. It does not matter which individual gold coin we draw into.
Thus there are only 2 possibilities, not 3, because we don’t care about draw order at all.
Bag 1 has a 100% chance of having a second gold in it.
Bag 2 has a 0% chance of having a second gold in it.
Which gives you a 50/50 chance of getting the gold.
Think of it this way, what the problem is really asking is, if I drew a gold coin, what are the odds that I have bag 1?



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Random, the fact that you drew a gold piece is given, you are factoring that into your probability calculation.
It is interesting how problems like these can boil down to assumptions.
Edit: what terrible toaster said



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which gold coin you pick in bag 1 actually does matter
there are two chances of bag 1 being picked and one chance of bag 2 being picked
look up bertrand’s box paradox



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Actually you are right, I’ve been framing my observation after I drew the first gold coin, when you are suppose to be framing it before (shame on me). Because I did it that way I avoid the consequence of the fact that drawing a gold coin has no ‘real’ effect on the probability of drawing GG, while it does effect the probabilities of GS and SS which, of course, makes my answer wrong.



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Again, in probability assumptions are everything.
The way Terrible toaster and I answered the problem is correct if the bag is essentially predetermined (given). Otherwise random’s approach is correct.
I think we all win.
