F.5 MATHS

Oh! So many questions, I think it is impossible to answer all, as it may exceed the limit set by the Yahoo system.

14a) The combination of getting 1 point :
(6,2), (6,3), (6,4), (2,6), (3,6), (4,6)
∴ the required probability is 6/36, ie. 1/6.

14b) The combination of getting 2 points :
(1,1), (1,2), (1,4), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (6,6)
∴ the required probability is 12/36, ie. 1/3

14c) The combination of getting more than 2 points :
(1,6), (5,6), (6,1), (6,5)
∴ the required probability is 4/36, ie. 1/9.

22ai) P(6,4) = 360

22aii) P(6,2) = 30

22b) P(n,4)
= n(n - 1)(n - 2)(n - 3)
= n(n - 3) (n - 1)(n - 2)
= (n^2 - 3n) (n^2 - 3n + 2)
= (n^2 - 3n)^2 + 2(n^2 - 3n)

22ci) P(m,4) >= 1680
(m^2 - 3m)^2 + 2(m^2 - 3m) - 1680 >= 0
(m^2 - 3m + 42)^2 - 3m - 40) >= 0
As it is known that x^2 - 3x + 42 is greater than zero for all x,
ie. m^2 - 3m + 42 is also always greater than zero.
∴ m^2 - 3m - 40 >= 0

22cii) m^2 - 3m - 40 >= 0
(m + 5)(m - 8) >= 0
m <= -5 (rej, as m >= 3) or m >= 8
∴ the possible range of m is : m >= 8

26ai) get 0 mark = 0.178
26aii) full mark = 0.000244
26aiii) pass the test = 0.0376

(All correct in part (a))

26bi) prob(2 students pass the test)
= prob(2 from have prepared) + prob(1 from yes, 1 from no) + prob(2 from no)
= C(30,2) / C(40,2) + 0.0376*C(10,1)*C(30,1) / C(40,2) + 0.0376^2*C(10,2) / C(40,2)
= 0.572 (corr to 3sf)

26bii) prob(at least 1 from do not prepared given that both pass the test)
= 1 - prob(2 from have prepared) / prob(both pass the test)
= 1 - [C(30,2) / C(40,2)] / (the result in part bi)
= 0.0254 (corr to 3 sf)

(Almost reach the Yahoo limit, sorry)