Year 12 Maths Exam Question

[HarryTiger]

NSW

Apparently it was very hard this year and they've given us one question from the exam.

Sample question
A game is played by throwing darts at a target. A player can choose to throw two or three darts.
Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice.
The probability that Darcy hits the target on any throw is p, where 0 < p < 1.
(i) Show that the probability that Darcy wins Game 1 is 2p - p[squared].
(ii) Show that the probaility that Darcy wins Game 2 is 3p[squared] - 2p[cubed].
(iii) Prove that Darcy is more likely to win Game 1 than Game 2.
(iv) Find the value of p for which Darcy is twice as likely to wine Game 1 as he is to win Game 2.

Read more: http://www.smh.com.au/nsw/hsc-stude...d-difficult-20111027-1mlg1.html#ixzz1byZkmAJk


I'm going to do part (i)

The probability of winning is 1 - Probability_Losing (easier this way as only one sequence loses).

= 1 - (miss in first try) * (miss in second try)
= 1 - (1 - p) * (1 - p)

expand the multiplication

= 1 - (1 - p - p + p^2)
= 1 - (1 - 2p + p^2)
= 1 - 1 + 2p - p^2
= 2p - p^2

 

[HarryTiger]

Question Time!

I meant to file this under Question & Answers where the smarties congregate. Pull out your abacus' conspiracy freaks and other associate nutters!

 

[The_Reaper]

ii) Hit first two times = p^2 (win)
Hit first time, missed second time, hit third time = p(1-p)(p) = p^2 - p^3 (win)
Hit first time, missed second time, missed third time = p(1-p)(1-p) = p-2p^2+p^3
Missed first time, hit second time, hit third time = (1-p)(p)(p) = p^2 - p^3 (win)
Missed first time, hit second time, missed third time= (1-p)p(1-p) = p-2p^2+p^3
Missed first time, missed second time, hit third time = (1-p)(1-p)(p) = p-2p^2+p^3
Missed all three times = (1-p)^3 = 1 - 3p + 3p^2 - p^3

Sum= 1

Probability = occurance/total
Sum of winning probabilities = P^2 + p^2 - P^3 + p^2 - p^3

Probablity = above divided by 1
= 3p^2 - 2p^3

 

[The_Reaper]

iii)
I have a very lame answer for this question which I am sure must somehow be wrong

2p-P^2>3p^2-2p^3

2p+2p^2>-2p^3

Since p is greater than 0
this is true

 

[The_Reaper]

IV)
2p-p^2 = 6p^2 - 4p^3
2p-7p^2+4p^3 = 0
p = 1.3903, 0.3596 and 0

1.3903 is discounted because p <1
0 is discountd because p>0

therefore p=0.3596

 

[HarryTiger]

iii)
I have a very lame answer for this question which I am sure must somehow be wrong

2p-P^2>3p^2-2p^3

2p+2p^2>-2p^3

Since p is greater than 0
this is true

I like to see this one done better, must be able to further reduce it some way. I forget how we were taught.

Substitute for p?

2(0 < p < 1).... no.

or maybe? allow for ranges

[0..2] - [0..1] = [0..1] (not sure if notation is right)

as p approaches 1 win prob will approach 1 too, makes sense but doesn't answer the question. Needs to be for a given p.

Anyone?

 

[The_Reaper]

I like to see this one done better, must be able to further reduce it some way. I forget how we were taught.

Substitute for p?

2(0 < p < 1).... no.

or maybe? allow for ranges

[0..2] - [0..1] = [0..1] (not sure if notation is right)

as p approaches 1 win prob will approach 1 too, makes sense but doesn't answer the question. Needs to be for a given p.

Anyone?

Yeah, it's a really bad answer.

Tried some other techniques I knew off the top of my head (squeeze theorem etc) but realised that they were all about limits.

 

[the fly]

p(Game 1) - p(Game 2) = 2p-p^2-(3p^2-2p^3)

=2p-4p^2+2p^3
=2p(1-2p+p^2)
=2p(1-p)^2>0

 

[HarryTiger]

p(Game 1) - p(Game 2) = 2p-p^2-(3p^2-2p^3)

=2p-4p^2+2p^3
=2p(1-2p+p^2)
=2p(1-p)^2>0

thanks, that wasn't so hard just had know what to start with.

 

[powerful girl]

what language is this thread written in, I don't understand any of it.

 

[courtjester]

If Darcy can't hit a target with a dart every time he is a bit of a tardy.

 

[Kerrby]

If Darcy can't hit a target with a dart every time he is a bit of a tardy.

I hope he doesn't take lessons from Nahas then .