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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 yell
by escherplex 11y ago
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 11y agomy 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.