# Hannah and her sweets

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 $6/n$. Now there are five orange sweets left out of $n-1$, so the probability that the second sweet is orange, assuming that the first one is, is $5/(n-1)$. Therefore we need to solve $(6/n) \times (5/(n-1)) = 1/3$. Multiplying it out gives ${30 \over n(n-1)} = {1 \over 3}$

and we can easily rearrange to get $90 = n(n-1)$. So $n = 10$; there are 10 sweets. (I guarantee you that a bunch of students went straight for the quadratic formula at this point – but you don’t have to, it’s easy to find two consecutive numbers that multiply to 90.) According to the BBC, setters of the exam point out that this was supposed to be one of the more difficult questions, “targeted at students aiming for A and A* grades”.

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 $r$ and the number of blue(berries) is $b$, then we have ${r \over r+b} {r-1 \over r+b-1} + {b \over r+b} {b-1 \over r+b-1} = {1 \over 2};$

the first term is the probability of drawing red-red and the second is the probability of drawing blue-blue. We can rewrite as $2(r(r-1) + b(b-1)) = (r+b)(r+b-1)$

but that isn’t much of a help, to be honest. Solving for $r$ in terms of $b$ gives $r = {1 \over 2} \left( 2b \pm \sqrt{8b+1} + 1 \right)$
and if you happen to know the obscure piece of trivia that for integer $b$, $8b+1$ is a perfect square if and only if $b$ is triangular, then you can show that $r$ and $b$ must be two consecutive triangular numbers. For example $b = 15$ leads to the solutions $r = 21$ and $r = 10$.

But if you use n = r + b, then is as pointed out at the Quora answer, $r = (n \pm \sqrt{n})/2$ (with different notation) and this doesn’t require knowing anything obscure. Setting $n = 36$ leads immediately to the solutions $r = 21, r = 15$, for example. In general, to make this work out the number of sweets needs to be a perfect square.

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?

## 6 thoughts on “Hannah and her sweets”

1. blaisepascal2014 says:

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).

1. Michael Lugo says:

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.

1. blaisepascal2014 says:

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.

2. statsinthewild says:

Reblogged this on Stats in the Wild.

3. Belinda Wong says:

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.

1. John Armstrong says:

“Without replacement”. The explanation is in the post.