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In 2015 this GCSE maths question caused many to say it was "too difficult" and demanding reduced pass marks:

"There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0."

I remember being pretty unimpressed. I'm sure my peer group could have answered it by age 12-13. I didn't shine at maths and hadn't looked at a text book in 50 years but still solved it in under a couple of minutes.
 
I found integration a mystery because nobody told me you have to guess what the answer might be and then verify the guess by differentiating it. Mind you, that wouldn't have helped to make the guess in the first place.

alanofcleeves said:
There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0.
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I must say that, at first glance, the connection between the story and "show that n²-n-90=0" is not obvious.
 
OK, so let me work through this (typing as I go, so this is my actual thought train).
alanofcleeves said:
There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0.
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The probability that the first sweet was orange is 6/n. The probability that the second sweet is also orange is 5/(n-1), so the total probability is 6*5/(n*(n-1)) = 30/(n²-n)

We are told the total probability is 1/3, so 1/3 = 30/(n²-n) or n²-n = 90 or n²-n-90 = 0
QED, but if that's "obvious once you've seen it" you're brighter than me.

The solution is obviously n=10.

Were the students then required to calculate n? If not, why not?
 
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Black Hole said:
Were the students then required to calculate n? If not, why not?
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As far as I know there was no requirement to solve the quadratic or to determine, say, how many 'yellow' sweets were in the bag.
Might have changed the question from being "too difficult" to "much too difficult".
 
I can't remember when we learnt to solve quadratics, possibly 2nd year at grammar school?
 
Black Hole said:
I found integration a mystery because nobody told me you have to guess what the answer might be and then verify the guess by differentiating it. Mind you, that wouldn't have helped to make the guess in the first place.
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Our first glimpse of uncertainty in maths. No infallible algorithm, just a few heuristics that might or might not work.
Proving what you said is no easy thing, It used to be the subject of 3rd year Uni courses, but is now 4th year or beyond.
 
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alanofcleeves said:
As far as I know there was no requirement to solve the quadratic or to determine, say, how many 'yellow' sweets were in the bag.
Might have changed the question from being "too difficult" to "much too difficult".
Click to expand...
That would be a separate question these days, ending

Given that n²-n-90=0, show that n=10.

Ie, (a) dont make them derive the quadratic, but (b) tell them what the answer was. It would be in A level.
 
One of my Scratlings decided to study Discrete for one of her A level modules. There was nobody at school to teach it, so she got permission to study it from a book, with my backup.

We had 5 contact hours, one for each chapter. That was a third of y11, or 2nd year 6th, on that A level. She passed easily. Plus, I found a huge mistake early on, in a definition in the book. This would affect the ability of any students to understand anything that followed.
 
Scrat said:
Yeah, as in finite, combinatorial, as opposed to continuous. Mainly graph theory, as in vertices and edges, max flow min cut, Dijkstra algorithm, spanning trees, etc.
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:confused:

Scrat said:
Dijkstra algorithm
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Had to look that up - I presume it's something like the travelling salesman problem (or another name for it).

Some time ago (and I might try again) I tried to get my head around elliptic curve cryptography (what elliptic curves have to do with ellipses is a mystery) - as these weird things seem to pass through integral points on the plane does that count as discrete?
 
By the way, I seem not to have seen much of the snooker - I'm blaming you and AoC for that (just printing the mathematical pi thing).
 
Black Hole said:
:confused:
(what elliptic curves have to do with ellipses is a mystery) -
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"Elliptic curves are not ellipses. The name comes from certain integrals involved in computing the arc length of an ellipse, which involve square roots of cubic and quartic polynomials in x."
 
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