ruse 2013 trial solutions (1 Viewer)

[aDimitri]

does anybody have a copy of solutions to the JRAH 2013 MX1 Trial? bloody tough paper for an MX1, need a copy of them soln's
much appreciated in advance

 

[faisalabdul16]

Such a good paper! First multiple choice and its a projectile question hahah although an easy one, just the thought of it being first question in paper lol i need the solutions too

 

[aDimitri]

i finished a timed attempt about an hour ago and i'd like to mark it :/

 

[faisalabdul16]

What were your answers for the multiple choice?

 

[Kurosaki]

Could you post the exam? We could all have a crack at solving it

 

[faisalabdul16]

Here is the paper

 

[aDimitri]

My multiple choice was
1- A
2- A
3- B
4- D
5- B
6- ?
7- C
8- D
9- A
10- ?

 

[faisalabdul16]

i got the same as you for them all, and i did not get the same questions as you. For question 10, i did 1- (none from James ruse being picked) and i got a quadratic...

 

[Kurosaki]

i got the same as you for them all, and i did not get the same questions as you. For question 10, i did 1- (none from James ruse being picked) and i got a quadratic...

that fails because that doesn't eliminate the possibility of there being only 1 Ruse student. I think you need knowledge of conditional probability to do this.

 

[aDimitri]

i didn't get the fact that it wasn't just (n-1)/(N-1)...

 

[Kurosaki]

i didn't get the fact that it wasn't just (n-1)/(N-1)...

That would be true for a slightly reworded problem; however, you have to take into account that you KNOW you have at least one James Ruse student. So you need to divide by the probability of having at least 1 ruse student (essentially restricting your sample space)

 

[iStudent]

For the last MCQ, I got A as the answer

What I did:

P = n(2ruse)/[n(1 ruse, 1 other) + n(2 ruse)]
= (nC2)/[nC1*(N-n)C1 + nC2]
= n(n-1)/2 / [n(N-n) + n(n-1)/2]
= (n-1)/2N-n-1
which is A

 

[aDimitri]

That would be true for a slightly reworded problem; however, you have to take into account that you KNOW you have at least one James Ruse student. So you need to divide by the probability of having at least 1 ruse student (essentially restricting your sample space)

See i completely disagree with this, if it says you KNOW you have at least one James Ruse student, it means the probability of having at least 1 James Ruse student is 1. So you're dividing by 1. It implies that the first case has already happened, otherwise it wouldn't even need to bother saying that you know there is one Ruse student, it could just say find the probability that they are both from James Ruse.

 

[JRAHS14]

My multiple choice was
1- A
2- A
3- B
4- D
5- B
6- D
7- C
8- D
9- B
10- A

 

[aDimitri]

9 is A.

 

[braintic]

9 is A.

Nup, 9 is B

 

[braintic]

See i completely disagree with this, if it says you KNOW you have at least one James Ruse student, it means the probability of having at least 1 James Ruse student is 1. So you're dividing by 1. It implies that the first case has already happened, otherwise it wouldn't even need to bother saying that you know there is one Ruse student, it could just say find the probability that they are both from James Ruse.

And no again.

Say the question was: You roll 2 dice, and you KNOW that at least one of the dice shows a 1, what is the probability that both show a 1.

There is only one (1,1) outcome. There are 11 outcomes showing a 1. So the answer is 1/11.

This is exactly the same as (1/36) / (11/36),
which is P(1,1) / P(at least one 1)

 

[aDimitri]

And no again.

Say the question was: You roll 2 dice, and you KNOW that at least one of the dice shows a 1, what is the probability that both show a 1.

There is only one (1,1) outcome. There are 11 outcomes showing a 1. So the answer is 1/11.

This is exactly the same as (1/36) / (11/36),
which is P(1,1) / P(at least one 1)

Yeah i already understood this lol i'm just saying that that's not at all how i interpreted that wording. Like i interpreted that as we know one dice is a 1, therefore there is a 1 in 6 chance of the other dice also being a 1.

Ah i now see why 9 isn't A, i assumed that closing the triangle passed through the center now i see it obviously doesn't.

 

[braintic]

Yeah i already understood this lol i'm just saying that that's not at all how i interpreted that wording. Like i interpreted that as we know one dice is a 1, therefore there is a 1 in 6 chance of the other dice also being a 1.

Ah i now see why 9 isn't A, i assumed that closing the triangle passed through the center now i see it obviously doesn't.

The probability question in the 1997 2 unit HSC had a similar issue, except that the wording meant that is WAS supposed to be interpreted in the way you did. Problem was the markers got it wrong, and everyone's paper was marked incorrectly. It took 10 years for MANSW to officially correct the answer.

 

[cutemouse]

Lol Q13c is basically copied from a Victorian paper ...

Q12c is very similar to a past HSC paper (2004 I think)

I've seen Q14a previously.

Other questions seem pretty standard as well.

Seems very unlike a typical Ruse paper. Perhaps the person writing it couldn't be bothered with spending too much time on it...

(and I'm not trolling)