# Leave poor Hannah alone!

Gosh, who would have thought it? 15,000 students (at the time of writing) have signed a petition to get EdExcel to change the grade boundaries for Thursday's non-calculator GCSE paper.

Good luck with that. The boundaries will have been set using magic statistics, and either have been or will be scrutinised by OfQual.

This blog isn't about grade boundaries, or (for that matter) about entitled whining. It's about Hannah and her sweets, a question that seems to have become the lightning rod for complaints.

My position: Hannah's sweets question was perfectly fair, and1 I'd go so far as to call it a good question.

### Background

The question (I paraphrase) was:

Hannah has 6 green sweets out of a total of $n$. If she picks one sweet and then another (without replacement), the probability of getting two green sweets is $\frac 13$. Show that $n^2 -n - 90 = 0$.

This is question in the high teens/early twenties, so about 80% of the way through the paper; if I were in the habit of attaching grades to questions, I'd put it towards the top end of the A-grade questions, so it's not supposed to be a walk in the park.

Quite on the contrary: these are meant to be the questions that differentiate competent mathematicians from good ones -- the questions that decide whether you go on to do A-levels in subjects that involve mathematical thinking.

### Solving it

Had the question given Hannah, say, 6 green sweets out of 15 and asked the probability of getting two green sweets, a student expecting an A would almost certainly work out $\frac{6}{15}\times \frac{5}{14} = \frac{1}{7}$ without any serious problems.

This is the same thing, only the thing you don't know is the '15'. You know that's $n$ -- so do sums with $n$!

The probability of getting two green sweets is $\frac {6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}$. You're told that equals $\frac{1}{3}$; multiplying by both denominators gives $90 = n(n-1)$, which expands to the equation you're asked to find.

### Why it's a good question

That jump from 'working with explicit numbers' to 'working with unknown numbers' is, I would say, one of the key indicators of whether a student is cut out for A-level. It's not (as some students have grumbled) an A-level question -- to me, it's quite a natural combination of algebraic fractions2 and probability3 and I'm surprised there aren't more questions of this nature in GCSEs.

The only one I've been able to find, though, was from the 2004 paper, which I've put here for the sake of comparison.

Edited 2015-06-06 to change balls back to sweets. Thanks to @MeganGuinan1 for the correction. ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. pseudocontext aside []
2. which you revised the hell out of, right? []
3. which you knew was going to come up []

### 14 comments on “Leave poor Hannah alone!”

• ##### Barney Maunder-Taylor

I agree, Colin. I’d also add that the part b of the same question then asked you to solve the quadratic equation that they’d already given you in part a – meaning that you could still attempt part b if you’d been unable to get part a. And it’s a nice example of a quadratic equation cropping up in a “real world” problem.

• ##### Colin

… for rather contrived values of ‘real world’ 😉

• ##### gav_cooper

the 2004 question was on the paper I sat! thanks for this little blast from the past! I don’t think I got the quadratic out in the end back then, haha.

• ##### Rob Brown

You said it much better than me. I couldn’t agree more. The only problem with this question is that neither the students nor the teachers saw it coming. Well done edexcel let’s hope we get more of the same in the new curriculum. I would say information about the yellow sweet was somewhat misleading but if any of my kids had complained (they didn’t) I would have told them to suck it up.

• ##### Colin

Thanks, Rob!

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