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yes, exactly that

got that :

C(I) that was just probability: P=(1/n^k-1){k(n-1)/(n-k)x(n-2)/(n-k-1)x...(n-k)/(...)} then simplify C(II) substitition of k+1 in I, then solve simultaneous (III) look for discrimant (iv)??

Hello>

Re: General Thoughts: Mathematics Ext 2 Rushed that question, i think i got (ar^2)/3. Polynomial was the hardest Q, absolutely no clues.