Mathematical Methods CAS Exam 2 Discussion - How was it??

[tony3272]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

[BoredSatan]

The exam was probably one of the best exams written by VCAA.

think about it.

the MC, filled with some challenging yet some easier problems

SA: A lot of parts didnt require working from the previous question, or they gave you the answers so you could work out the rest.

I mean you can never expect that methods was going to be REALLY easy, if it was then VCAA would have the hardest time figuring out how to allocate marks, and how to seperate the best from the rest.

[geoff_821]

SO much more happy with this exam then the last one.
Apart from the last question and about 3 multiple choice questions, everything was pretty good.
Hopefully scrape a few method marks on the last questions

[ttn]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

[jamesh]

how did you guys solve 2e

[luffy]

I just did the exam and I'll give a bit of a comparison with our exam last year.

Length: I thought our exam was slightly longer, but that was likely because I did ours under that extra pressure.

Difficulty: This is up for debate. I thought they were roughly equal. Ours wasn't too bad with the exception of our last question - the same applies to this years exam. The multiple choice was roughly the same difficulty as ours as well - perhaps slightly harder.

Though, overall, I found ours harder (likely because of the pressure in my one). Hence, I think they were roughly equal in difficulty. The last part of the last question really does make you think, just as ours did last year.
For those who are curious, I haven't checked my solution, but my answer was . If I lost a mark, it would be on that question. The reason it is greater than or equal is because if k increases, the time taken to swim increases. Thus, the minimum time would be when you go directly towards the plant.

I'll scan the exam later for the rest of you guys.

[BoredSatan]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

i dont think that works because then the denomintor is greater than 0

it should be Pr(X>3)*0.5/[Pr(X>3)*0.5+Pr(Y>3)*0.5]=0.4103

you have to times 0.5 because there were equal numbers of A and B chocolates in the bag/tub/whateever

[HCbigstick]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

+1. I did the same thing and got 26/49 which was 0.5306 to 4 dp. Anyone else?

[BoredSatan]

I just did the exam and I'll give a bit of a comparison with our exam last year.

Length: I thought our exam was slightly longer, but that was likely because I did ours under that extra pressure.

Difficulty: This is up for debate. I thought they were roughly equal. Ours wasn't too bad with the exception of our last question - the same applies to this years exam. The multiple choice was roughly the same difficulty as ours as well - perhaps slightly harder.

Though, overall, I found ours harder (likely because of the pressure in my one). Hence, I think they were roughly equal in difficulty. The last question really does make you think, just as our last question did.
For those who are curious, I haven't checked my solution, but my answer was . If I lost a mark, it would be on that question. The reason it is greater than or equal is because if k increases, the time taken to swim increases. Thus, the minimum time would be when you go directly towards the plant.

I'll scan the exam later for the rest of you guys.

he full marked exam 2 last year. he cant be wrong. but that means IM wrong

[ttn]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

+1. I did the same thing and got 26/49 which was 0.5306 to 4 dp. Anyone else?

I got 0.4103 like the others?

[HCbigstick]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

i dont think that works because then the denomintor is greater than 0

it should be Pr(X>3)*0.5/[Pr(X>3)*0.5+Pr(Y>3)*0.5]=0.4103

you have to times 0.5 because there were equal numbers of A and B chocolates in the bag/tub/whateever

My denominator was 0.9423 when I did that?

[ttn]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

i dont think that works because then the denomintor is greater than 0

it should be Pr(X>3)*0.5/[Pr(X>3)*0.5+Pr(Y>3)*0.5]=0.4103

you have to times 0.5 because there were equal numbers of A and B chocolates in the bag/tub/whateever

Your 0.5s just cancels out anyway?

0.5*Pr(x>3)
--------------------------------
0.5*[ Pr(x>3) + Pr(Y>3) ]

[HCbigstick]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

+1. I did the same thing and got 26/49 which was 0.5306 to 4 dp. Anyone else?

I got 0.4103 like the others?

Must have used the wrong value for Pr(Y>3). Hopefully that's only one mark lost.

[BoredSatan]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

i dont think that works because then the denomintor is greater than 0

it should be Pr(X>3)*0.5/[Pr(X>3)*0.5+Pr(Y>3)*0.5]=0.4103

you have to times 0.5 because there were equal numbers of A and B chocolates in the bag/tub/whateever

My denominator was 0.9423 when I did that?

Pr(X>3) is 0.5 beacuse the mean is 3
Pr (Y>3) is 1-(9/32)=23/32
if you add them you get greater than 1..

EDIT: I guess you can do it that way too..

[acinod]

Did anyone get 0.4103 for part 2e? I'm not sure about that one.

YESSSSSSSSS I was quite uncertain about that one

Lol i got that. What method did people use to solve it? Mine was weird.

I did

Pr(X>3) / [ Pr(X>3) + Pr(Y>3) ]

+1. I did the same thing and got 26/49 which was 0.5306 to 4 dp. Anyone else?

Pr(Chocolate A|Dark Chocolate)=Pr(Dark Chocolate A)/Pr(Dark Chocolate)=Pr(X>3)/(Pr(X>3)+Pr(Y>3))=0.5/(0.5+23/32)=0.4103