2017 Methods Exam 1 [Solutions]

[Jiggleling]

I don't think it'll be scratched but they would accept both answers such as if you stated tangent is undefined due to endpoint and intersection of both points, although I might be wrong in my understanding

[Steve252]

For the last question, how many marks out of 4 would we get for just the equation of AC?
I got the equation but then equated it to a wrong equation by accident so I'm only gonna get marks for the equation of AC

[Syndicate]

For the last question, how many marks out of 4 would we get for just the equation of AC?
I got the equation but then equated it to a wrong equation by accident so I'm only gonna get marks for the equation of AC

You should get 3 marks, given that you got the right equation for AC, and your working out is all correct. You would only lose a mark for getting the wrong coordinates.

[Steve252]

Really, that would be amazing. Yep I've confirmed that my equation for AC was correct, with proper working out as well
Thanks so much

[Rieko Ioane]

There's a good chance they could have 1 mark for the correct x-coord and 1 mark for the correct y-coord fr

[finn_smith]

Hi guys, a question that seems to be highly contentious at my school, a relatively strong cohort, is whether the final question of the Computer Login probability section was conditional? Many argue, including myself, that because it states "Jac logs on to the computer successfully", that it implies it is a conditional question, very similar to that of one of the questions in the 2012 VCAA Exam, where there was another hidden conditional probability question that only 3% of the state got full marks for (according to the official VCAA Examination Report). So, what we were proposing was that the value obtained from the previous question, which was asking for him to successfully login in entirely, (which was 98/125) would be used in the final question. We saw it as a conditional by saying find the probability that he logs in on the 2nd or 3rd attempt, given that he logs in to the computer successfully (found to be 98/125 in the previous question), as it was stated in this question that 'Jac logs on to the computer successfully". Hence, our method went on to have the answer as (2/5 x 3/5) + (3/5 x 3/5 x 2/5) / (98/125). This gave an answer of 24/49. Although I can also see why it may be 48/125, I believe it is a conditional question. Thanks

[Sine]

Hi guys, a question that seems to be highly contentious at my school, a relatively strong cohort, is whether the final question of the Computer Login probability section was conditional? Many argue, including myself, that because it states "Jac logs on to the computer successfully", that it implies it is a conditional question, very similar to that of one of the questions in the 2012 VCAA Exam, where there was another hidden conditional probability question that only 3% of the state got full marks for (according to the official VCAA Examination Report). So, what we were proposing was that the value obtained from the previous question, which was asking for him to successfully login in entirely, (which was 98/125) would be used in the final question. We saw it as a conditional by saying find the probability that he logs in on the 2nd or 3rd attempt, given that he logs in to the computer successfully (found to be 98/125 in the previous question), as it was stated in this question that 'Jac logs on to the computer successfully". Hence, our method went on to have the answer as (2/5 x 3/5) + (3/5 x 3/5 x 2/5) / (98/125). This gave an answer of 24/49. Although I can also see why it may be 48/125, I believe it is a conditional question. Thanks

hey this is a valid pretty valid question
2012 question was "daniel recieves telephone calls on both monday and tuesday What is the probability that Daniel receives a total of four calls over these two days?"
2017 question was  "Calculate the probability that he logs on to the computer successfully on the 2nd or the third attempt"
 so some subtle differences
Hard to explain reasoning other than intuition
itute also has their answers up and they got 48/125

[finn_smith]

hey this is a valid pretty valid question
2012 question was "daniel recieves telephone calls on both monday and tuesday What is the probability that Daniel receives a total of four calls over these two days?"
2017 question was  "Calculate the probability that he logs on to the computer successfully on the 2nd or the third attempt"
 so some subtle differences
Hard to explain reasoning other than intuition
itute also has their answers up and they got 48/125

Thanks for this response! But what also may be subtle hint to further push my case is that the previous question asked "in the form of a/b" being 98/125. With the next question asking about 2nd or 3rd attempt, the question asked "in the form of c/d", meaning they used different constants, instead of saying on "c/b". Although this may essentially give away the denominator's value if it were to be this way (where b = 125), I have seen them use on practice exams the repetition of a constant if it is applicable in multiple stems within a question. Food for thought

[Sine]

Thanks for this response! But what also may be subtle hint to further push my case is that the previous question asked "in the form of a/b" being 98/125. With the next question asking about 2nd or 3rd attempt, the question asked "in the form of c/d", meaning they used different constants, instead of saying on "c/b". Although this may essentially give away the denominator's value if it were to be this way (where b = 125), I have seen them use on practice exams the repetition of a constant if it is applicable in multiple stems within a question. Food for thought

two constants can be the same
consider a line in the form y= mx + c it is possible to have a line such that m = c so y =x + 1

[jazzycab]

Hi guys, a question that seems to be highly contentious at my school, a relatively strong cohort, is whether the final question of the Computer Login probability section was conditional? Many argue, including myself, that because it states "Jac logs on to the computer successfully", that it implies it is a conditional question, very similar to that of one of the questions in the 2012 VCAA Exam, where there was another hidden conditional probability question that only 3% of the state got full marks for (according to the official VCAA Examination Report). So, what we were proposing was that the value obtained from the previous question, which was asking for him to successfully login in entirely, (which was 98/125) would be used in the final question. We saw it as a conditional by saying find the probability that he logs in on the 2nd or 3rd attempt, given that he logs in to the computer successfully (found to be 98/125 in the previous question), as it was stated in this question that 'Jac logs on to the computer successfully". Hence, our method went on to have the answer as (2/5 x 3/5) + (3/5 x 3/5 x 2/5) / (98/125). This gave an answer of 24/49. Although I can also see why it may be 48/125, I believe it is a conditional question. Thanks

Unfortunately for you, this won't be the case. We need to consider whether the universal set has indeed been reduced.
In this entire question, there are 4 total possibilities:
Jac logs in on the 1st attempt
Jac logs in on the 2nd attempt
Jac logs in on the 3rd attempt
Jac fails to log in
There is nothing in this question to indicate that none of these are possible any more.
Consider a similar scenario:

A plays B in a best of 5 sets game of tennis.
Find the probability that A beats B in 3 or 4 sets.

A simplified universal set is:
{A wins in 3 sets, A wins in 4 sets, A wins in 5 sets, A does not win}

The probability that we are looking for is obtained by finding:
Pr(A wins in 3 sets) + Pr(A wins in 4 sets)/(Pr(A wins in 3 sets) + Pr(A wins in 4 sets) + Pr(A wins in 5 sets) + Pr(A loses))
= Pr(A wins in 3 sets) + Pr (A wins in 4 sets)

The bolded question above asks us to find the probability that A beats B in some way, but doesn't indicate that A DEFINITELY beats B, hence there is no condition to consider.

[finn_smith]

Unfortunately for you, this won't be the case. We need to consider whether the universal set has indeed been reduced.
In this entire question, there are 4 total possibilities:
Jac logs in on the 1st attempt
Jac logs in on the 2nd attempt
Jac logs in on the 3rd attempt
Jac fails to log in
There is nothing in this question to indicate that none of these are possible any more.
Consider a similar scenario:

A plays B in a best of 5 sets game of tennis.
Find the probability that A beats B in 3 or 4 sets.

A simplified universal set is:
{A wins in 3 sets, A wins in 4 sets, A wins in 5 sets, A does not win}

The probability that we are looking for is obtained by finding:
Pr(A wins in 3 sets) + Pr(A wins in 4 sets)/(Pr(A wins in 3 sets) + Pr(A wins in 4 sets) + Pr(A wins in 5 sets) + Pr(A loses))
= Pr(A wins in 3 sets) + Pr (A wins in 4 sets)

The bolded question above asks us to find the probability that A beats B in some way, but doesn't indicate that A DEFINITELY beats B, hence there is no condition to consider.

‘Jac fails to log in’ can be ruled out, as the question states ‘he successfully logs in’ - hence my reasoning why the set is reduced from 125/125 to 98/125? Or is there a better way you could explain why this isn’t the case? I’m just asking out of genuine curiosity now

[Sine]

good questions

It would be conditional if the question would say something like "Jac sucessfully logs in...find the probability that is on the 2nd or 3rd attempt"
but the question says "calculate the probability that he logs on to the computer successfully on the 2nd or the third attempt"

the quesiton merely asks for the PROBABILITY that it occurs and probaility is inherently = (success/ success + failures)

[jazzycab]

‘Jac fails to log in’ can be ruled out, as the question states ‘he successfully logs in’ - hence my reasoning why the set is reduced from 125/125 to 98/125? Or is there a better way you could explain why this isn’t the case? I’m just asking out of genuine curiosity now

Further to Sine's comment, if you compare to the other example I gave above, my question stated "Find the probability that A beats B..." however, this isn't an indication that A does indeed beat B.

I guess one other way to think about it is to try and describe, qualitatively, the probability required in a different way.
If we want to find the probability of Jac logging on successfully on the 2nd or 3rd attempt out of the entirety of the universal set, is there a different way that we could describe it?
I can't think of anything better or more concise than what was asked on the exam, however, it is quite easy to think of a way of rephrasing it to force a conditional (by making this condition fairly obvious, similar to Sine's description above).