Apparently students in the UK have been protesting against the following question on a GCSE math exam (see e. g. coverage at The Guardian):
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.
The probability that the first sweet is orange is 
and we can easily rearrange to get 
There’s a question this reminds me of, of which I don’t recall the original source: there are r raspberry sweets and b blueberry sweets in a bag. You take two of them at random; the probability that they have the same flavor is exactly 1/2. What are possible values for r and b? (Okay, so I’ve heard it with “red” and “blue”, but let’s go with fruit flavors.) (Okay, so I’ve heard it with “red” and “blue”, but let’s go with fruit flavors.) See for example this Quora question. This one is a bit trickier, and depends on getting lucky and choosing the right parametrization of the problem. If the number of reds/raspberries is 
the first term is the probability of drawing red-red and the second is the probability of drawing blue-blue. We can rewrite as
but that isn’t much of a help, to be honest. Solving for
and if you happen to know the obscure piece of trivia that for integer 
But if you use n = r + b, then is as pointed out at the Quora answer, 
What if there are three colors of sweets? How can we choose the number of sweets of each color to make the probability of getting a match equal to one-third?
God plays dice 
To me, the problem with the original problem is that the problem isn’t “Find n” (which you do quite simply), it’s that the problem is “Show that n^2-n-90 = 0”. You found n, you didn’t show that the quadratic equation is true (and in fact, you comments on students going to the quadratic formula, which is only “necessary” if you are trying to solve for n).
That’s a good point. I’ve seen photos like the one that Alex Bellos posted. This photo says “(a) Show n^2-n-90 = 0”; the (a) seems to imply that there’s a (b), which I’d have to guess is “find n”. But this is getting lost in the uproar.
I don’t know any reason to develop that quadratic equation on the way to “find n” except if you were intended to do rote invocation of the quadratic formula. In which case, I found it helpful to know there are 361 points on a goban, but I don’t expect that to be common knowledge.
As you pointed out, there’s no reason to use the QF for this problem. Heck, even the method of factoring x^2+ax+b by finding (p,q) such that p+q=a and pq=b (or, p+q=-1, pq=-90) is easier than the QF, and is functionally what you did.
Reblogged this on Stats in the Wild.
Its a bit late to be commenting on Hannah’s sweets, but for what its worth, here’s my comment. The probability of any sweet taken from the bag being orange is 6/n. That applies to each and every one of them. The probability of any two of them both being orange is 36/(n x n). The probability can never be 1/3, its impossible. The examiners got it wrong.
“Without replacement”. The explanation is in the post.