[escherplex]

Not if this is considered a particularly difficult UK Maths GCSE exam question, which over 90 percent couldn't answer:

Hannah has 6 orange sweets and some yellow sweets. Overall, she has n sweets. The probability of her taking 2 orange sweets is 1/3. Prove that: n^2-n-90=0.

If a HS student I would imagine you would

First: think of coins p(H1) = .5; p(H2) =.5; p(H1+H2) = .5 * .5 = .25

Second: OK, here (6/n) * (5/(n-1)) = 1/3

Third solve: 30/(n * (n-1)) = 1/3

                     90 = n^2 - n
                      0 = n^2 - n - 90

n=10 or -9; if -9 then Hannah only has one sweet and lifted 9 from somebody else just to demo her point.

(from London Telegraph today)

[thaumasiotes]

my instinct for approaching the problem:

    C(6,2) / C(n,2) = 1/3

(where C(n,k) is n choose k)

    15 / [n(n-1)/2] = 1/3
    30 / n(n-1) = 1/3
    n(n-1) = 90

I probably wouldn't have done that in high school, though.