Do this maths. It R Hard.
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And had (b) and onwards been included, I'd have known what else they were looking for, and would have had more information to select a suitable approach for (a).
You did not need any more information, just a bit of basic reading comprehension.
They present an argument for why the probability is only true with one n (at the end), and find that the equation holds for that single value. Proof by exhaustion.
The point is that they didn't do that to start with and still thought that they'd answered the damn question.
I agreed that when RacePro eventually added the last part, it was a correct solution. That doesn't contradict me at all.
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I'm not going through all this shit again.
I answered the question in full as presented. End of.
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I'm not going through all this shit again.
I answered the question in full as presented. End of.
Fine. But it won't stop you having been totally wrong to start with and persisted in being wrong for some time (and I'm still not convinced that you understand what was wrong and why the last part you added was necessary, but whatever), and it won't make the amount of information given lacking in any way.
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That's actually a bastardization of notation. (o0 o1) "means" the probability of "grabbing an orange first, and an orange second". Which is the probability of "grabbing an orange first" (o0) times the probability of "grabbing an orange second after grabbing an orange first"(o1).
I'd like to know what you think it means, though :P
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Um I know exactly what the notation means. And it's still true that
P(o0 o1) != P(o0)P(o1)in general
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Oh, what does o0 and o1 mean when you write it?
And what does writing P(x y) mean?
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P(AB)is the probability of the eventAB. Which, in general, is notP(A)P(B). In general, it is
P(A)P(B|A).P(A)P(B|A)is, general, only equal toP(A)P(B)whenAandBare independent events. But the probability thatBoccurs depends on whetherAoccurred first.
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As I said, I misused notation because I didn't recall the right notation.
You're complaining that my notation is wrong and doesn't apply in the general case, I'm telling you I didn't use the notation you are using. I used my own "special" notation, which works in this case.
To translate: if o0 is A, and o1 is B, o0 o1 is C, not AB.
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You can solve the original problem otherwise (for example, prove that as the number of sweets grows, the probability drops, then basically do a binary search). In this case, you solve the problem, and then prove that for each solution the equation holds.
Or you can solve the problem the intended way, solve the quadratic, then test the answers. You did the same work twice (since it's already proven that the equation holds at the moment it shows up in your calculations for the original problem), but it's still a valid proof.
At no point do you need to assume that the equation holds.
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prob(o0 o1) = 1/3 = p(o0) * p(o1)
Objection! The equality
p(o0 o1) = p(o0) p(o1)only holds if o0 and o1 are independent. That is clearly not the case here, as the result of the first draw affects the probabilities of the second.Also, starting subscripts at 0 here is weird. Mathematicians start at 1.
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Objection! As I already explained to Captain, I'm not using lame math notation. I'm using my own ad-hoc notation. Note I wrote "prob(o0 o1)", not "p(o0 o1)". Show me one math textbook where they call the probability of something "prob".
Also, as I explained to Captain already, (o0 o1) is not o0 followed by o1, it is a separate symbol in its own right.
Since I'm not writing math notation, I'm free to use whatever subscript I want, and I want to start at 0.
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As we've explained, the events are
o0 - draw an orange sweet from a bag of n sweets, including 6 orange ones
o1 - draw an orange sweet from a bag of n-1 sweets, including 5 orange onesThe notation still sucks, but this kinda makes it work.
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Macie's explanation is nicer than mine, so I'm going to go with it.
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If you're making up your own notation and not explaining it before using it then I can't accept your proof.
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Shush, you pedant, he's right. And the way of getting to n^2 - n - 90 = 0 is largely irrelevant to the proof, and the correct way is contained above.
The point is - you can do the proof by exhaustion.
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Well, I could go back and translate to standard notation. Do I win anything if you accept my proof to make it worth my time?
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The point is - you can do the proof by exhaustion.
How, exactly, do you plan on exhausting the natural numbers?
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Respect.
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I think the answer to that question should be obvious.
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You can prove that as the number of sweets goes up, the probability goes down in N.
Then you say "okay, so 10 gives 1/3, and 11 gives less, and 9, gives more, ergo 10 is the only solution in the domain of N".
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Prove there's an upper limit above which there can not be solutions (or, alternatively, above which every number is a solution).
Then from there the rest is trivial - just calculate the limit, and test all numbers below that limit.
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I like your explanation better. More generality, and it's an actual proof technique to remember.
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Let p(A) be the probability of grabbing two orange sweets from a bag initially containing n sweets, 6 of which are orange.
p(A) = 1/3
Let p(B) be the probability of grabbing one orange sweet from a bag initially containing n sweets. 6 of which are orange.
p(B) = 6 / n
Let p(C) be the probability of grabbing a second orange sweet from a bag initially containing n sweets, 6 of which are orange, after grabbing an orange sweet.
p(C) = (6-1) / (n-1)p(A) = p(B) * p(C) = 1/3
(6 / n) * (5 / (n-1)) = 1/3
(65) / (n(n-1)) = 1/3
30 / (n^2 - n) = 1/3
30 / (1/3) = n^2 - n
30 * 3 = (n^2 - n)
90 = n^2 - n
n^2 - n - 90 = 0
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Haw haw peer pressure.
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P(A)P(B|A) is, general, only equal to P(A)P(B) when A and B are independent events. But the probability that B occurs depends on whether A occurred first.
However, in this case we were really talking about
P(AB'), whereB'is equivalent toB|A. Which is to say, we're talking about the probability of picking an orange sweet a second time in the state where one was picked first. This is not the general probability of “the second sweet picked was orange” which the question made no mention of. [spoiler](With general random selection of sweets, the probability that the second sweet is orange is 0.6 (= 6/10*(6-1)/(10-1) + (10-6)/10*6/(10-1)). In case anyone gives a damn.)[/spoiler]
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Wtf has no one in this thread ever heard of guess-and-check?
@PleegWat said:
Factorization in the general case is trial and error. Remembering the formula for solving quadratic equations is meh.
QFTOh I see, it's ok when you do it, but it's not ok when @raceprouk does it?
Iterating values of n from 1 to 10 is way simpler than mucking around with PaPb and pre-calc and what not, and dismissing an intuitive, straightforward solution as ‘inelegant’ is totally unjustified.
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I think the most important thing to be learned from this discussion is that we'd be much better off teaching kids programming rather than algebra, because it is actually possible to allow unlimited creativity in problem solving while still having a standardized test result. One that can be produced without any human intervention, even.
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So there's nothing wrong with this fizzbuzz solution, as it prints the right results?
echo 1 echo 2 echo fizz echo 4 echo buzz ...Beauty is subjective, in maths as in programming.
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I think it was originally @Jaloopa that did it, not @RaceProUK.
There's nothing wrong with it in principle, mathematically. I expect most mathematicians won't bat an eyelid if you derive a quadratic or cubic equation and proceed to state the zeroes are A, B, and C without working it out explicitly.
However, I doubt the kids who originally got this question would get full marks for that approach.
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Which is exactly the point: a perfectly logical solution is not accepted here, because it doesn't match the solution the author of the test had in mind. This isn't a math question, it's a fucking psychology test.
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Everything is wrong with fizz buzz, don't you read my blog?
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I was the one who deducted there's only one possible solution, then had to defend in excruciating detail against arguments that weren't part of the original question, including the irrelevant detail that somehow the formula provided had to be shown to be the only formula that answered the question.
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I was the one who deducted there's only one possible solution,
You were the one who decided to answer something irrelephant.
The question, as stated, was not asking for the value of n but rather for the formula for calculating it and to demonstrate that it was the same as the formula given. Which follows from basic probability and formula simplification. There's probably a later step where the formula is then solved (with consideration of non-physical answers) — indeed, the structuring of the question shows that there's definitely at least one later step — but we're not dealing with that here.Did you never learn the cardinal rule of exams: answer the question and not something else?
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I answered the question as it was presented. If you think I did anything different, go shove it up your arse.
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I answered the question as it was presented.
You identified that ‘n = 10’ will satisfy the probability conditions and that it is value for which the equation is satisfied. You didn't show that the statement of the probability conditions was equivalent to the equation (plus a side condition that n ≥ 6). Equivalence is much stronger than co-satisfaction, as it illustrates why that equation was chosen instead of, say, n2 - 15n + 50 = 0.
Lots of maths just requires being a pendantic dickweed in one's rule-following. It's why computer-aided proof systems work so well…
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I showed the equivalence, and a lot more besides.
Fuck off.
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Oh I see, it's ok when you do it, but it's not ok when @raceprouk does it?
What the hell are you talking about?Iterating values of n from 1 to 10 is way simpler than mucking around with PaPb and pre-calc and what not, and dismissing an intuitive, straightforward solution as ‘inelegant’ is totally unjustified.
It's not intuitive, it's not straightforward, no-one did trial and error and if they had that wouldn't have been the issue with that 'solution'. And even the part that isn't a solution at all, and needs a horrible unintuitive, non-rigourous 'proof' that it's a biimplication to make it actually answer the question, is not simpler than 6 lines of simple construction.
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It's not intuitive, it's not straightforward, no-one did trial and error
That's what she's saying, though. There was no need to do trial and error because the correct guess is given to you right off the bat. Doesn't change the fact that plugging the answer back into the equation and showing that it works out is a perfectly cromulent way of showing that it is correct.and if they had that wouldn't have been the issue with that 'solution'. And even the part that isn't a solution at all, and needs a horrible unintuitive, non-rigourous 'proof' that it's a biimplication to make it actually answer the question, is not simpler than 6 lines of simple construction.
Except the part where you somehow managed to lose sight of the fact that there's only one possible value of n that could fit the given constraints.
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Except the part where you somehow managed to lose sight of the fact that there's only one possible value of n that could fit the given constraints.
Yes, but the question was essentially “where the fuck did that formula come from?” It certainly isn't one of the standard ones that are on the syllabus as tools to use for solving things (unlike, e.g., the quadratic root finder). Because that's what the question really is, asserting that you've found the value of n doesn't help solve it at all.
I find it fascinating how worked up some people seem to be getting over this. It's a trivial bit of maths; just writing down definitions (either on the syllabus or given in the question) and simplifying an equation. Standard stuff.
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Students need to look past what the question's actually asking to find the secret hidden bonus question inside?
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Does it ask for the value of n?
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Does it say you can't use the value of n in your solution?
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Does it say you can't use the value of n in your solution?
It also doesn't say you can't use a purple dragon dildo.
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The question, as stated, was not asking for the value of n but rather for the formula for calculating it and to demonstrate that it was the same as the formula given.
BZZZT!
It was asking to prove that all the solutions satisfy the formula. No more, no less. That the easiest way to do it is to solve the problem partway and arrive at the formula is irrelevant.
You didn't show that the statement of the probability conditions was equivalent to the equation
That's not what was asked for.
It also doesn't say you can't use a purple dragon dildo.
If you can derive a purple dragon dildo from the original problem, then use it to prove that all the solutions satisfy the formula, which is what's asked for, then I see no problem with that.
Filed under: purple dildo is a valid proof technique
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Except the part where you somehow managed to lose sight of the fact that there's only one possible value of n that could fit the given constraints.
Yes but that needed to be proved. It eventually was, more or less, and I agreed that that was, technically, a correct solution.
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Yep.
Show that n^2 - n - 90 = 0, where n is defined above as the number of sweets in the bag that makes the probability 1/3 (or in other words, as a solution to the problem).
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I remain fascinated that people object so much to using a fast, constructive method and instead insist on persist on using a slower, more complicated, non-constructive method.
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We know we can derive the formula from the problem without touching the solution, but that's not fun to argue.
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