Mathematics Extension 1 Marathon HSC 09 (2 Viewers)

[wonnny]

new question

i can do the first part, and step 1 of the induction, but then urgh - i know i gotta use part one :S
* oops thats meant to be cotΘ in the first part

 

[jet]

Now off to study for SOR
Someone else ask a q.

 

[Trebla]

Explain the flaw in this working.

 

[scardizzle]

(-1)^3 cannot equal (-1) ^6/2 as you are rooting a negative number

since (-1)^6/2 = ((-1)^1/2)^6

idk if that's right...

 

[cutemouse]

Explain the flaw in this working.

Line 1.

The fact that a=a^3 is not correct, where a is a real number (except for a=1).

You should write that (-1) = ((-1)^(1/3))(3) instead.

Also with Line 3. You need to take the square root first.

New Question:

Two children play a game which involves throwing a fair die once each. A win occurs if a player throws a 4, 5 or 6 and their opponent throws a 1, 2 or 3. A draw occurs if both players throw a 1, 2 or 3 or if both players throw a 4, 5 or 6.

(i) Find the probability of either player winning a game. [1 mark]

(ii) Find the probability that after 4 games at least half have been won by either of the players. [2 marks]

(iii) Find the probability that after n games, where n is an odd integer, at least half have been won by either of the players. Give reasons for your answer. [3 marks]

 

[scardizzle]

Line 1.

The fact that a=a^3 is not correct, where a is a real number (except for a=1).

You should write that (-1) = ((-1)^(1/3))(3) instead.

Also with Line 3. You need to take the square root first.

New Question:

Two children play a game which involves throwing a fair die once each. A win occurs if a player throws a 4, 5 or 6 and their opponent throws a 1, 2 or 3. A draw occurs if both players throw a 1, 2 or 3 or if both players throw a 4, 5 or 6.

(i) Find the probability of either player winning a game. [1 mark]

(ii) Find the probability that after 4 games at least half have been won by either of the players. [2 marks]

(iii) Find the probability that after n games, where n is an odd integer, at least half have been won by either of the players. Give reasons for your answer. [3 marks]

Are you sure you aren't mistaking a = a^3 with a = a^2?
as I see nothing wrong with the statement that -1 = -1^3

anyway...

im not sure if this is correct but here goes...

i) P(choose 1,2,3) = 1/2

P(choosing 4,5,6) = 1/2

P(player 1 winning) = 1/2 x 1/2 = 1/4

P(player 2 winning) = 1/2 x 1/2 = 1/4

therefore P(one win) = 1/4 + 1/4 = 1/2

ii) P(atleast half) = P(2 wins 2 ties) + P(3 wins 1 tie) + P (4 wins)

= (1/2)^2 x (1/2)^2 x 4C2 + (1/2)^3 x 1/2 x 4C3 + (1/2)^4

= 11/16(im sure there's an easier method)

iii) (0.5)^n [nC(n+1)/2 + nC(n+3)/2 + nC(n+5)/2 + ... + nC2n/2]

(cant think of a method to simplify this...

 

[cutemouse]

Perhaps what I should've said that a=a^3 is not necessarily correct. BTW parts (i) and (ii) are correct.

 

[taggs-sasuke]

HSC 2006 6 (a) (ii) L^2 = 2V^2(1-sin@)t^2 - 2aVcos@t + a^2 The right hand side of L^2 is a quadratic in t, which has a minimum value since 2V^2(1-sin@) > 0 for 0 < @ < pi/2. My question is how?