Official Exam 2 Discussion

[hanhanchampion]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

[kurrymuncher]

after checking the answers, i dropped 4 marks, oh shitshitshitshitshit!!!!!
If i get 38 for exam 1, is it possible for me to get a ss over 40 raw??

Yes it is.

[Synesthetic]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

Mean = integral over [2,6] of xf(x)dx, enter into calculator => 4.1333 hours

[hanhanchampion]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

Mean = integral over [2,6] of xf(x)dx, enter into calculator => 4.1333 hours

is that question 1 a iii?  i think it's the last question of 1......  i got that though...

[hanhanchampion]

after checking the answers, i dropped 4 marks, oh shitshitshitshitshit!!!!!
If i get 38 for exam 1, is it possible for me to get a ss over 40 raw??

Yes it is.

thank god im really scared cause of the competition!!

[Synesthetic]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

Mean = integral over [2,6] of xf(x)dx, enter into calculator => 4.1333 hours

is that question 1 a iii?  i think it's the last question of 1......  i got that though...

Sorry:

What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Combinatorics solution:

• Given exactly 6 successes == given exactly 2 failures
• Our probability space has these 2 failures anywhere in the 8 attempts, so 8 choices for the first fail * 7 choices for the second fail = 56
• The event we want has these 2 failures anywhere in the last 4 attempts, so 4 choices for the first fail * 3 choices for the second fail = 12
• 
  

[vipershadow]

so I've got at the very best, 56/80
Please tell me that could be a B+ : '(

[hanhanchampion]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

Mean = integral over [2,6] of xf(x)dx, enter into calculator => 4.1333 hours

is that question 1 a iii?  i think it's the last question of 1......  i got that though...

Sorry:

What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Combinatorics solution:

• Given exactly 6 successes == given exactly 2 failures
• Our probability space has these 2 failures anywhere in the 8 attempts, so 8 choices for the first fail * 7 choices for the second fail = 56
• The event we want has these 2 failures anywhere in the last 4 attempts, so 4 choices for the first fail * 3 choices for the second fail = 12
• 
  

Yes, I saw that, i just don't understand...  it seems to be explaining a lot..

[ReVeL]

so I've got at the very best, 56/80
Please tell me that could be a B+ : '(

Would have been an A last year. I would say this year was similar if not abit harder. So could very well be an A.

[Rietie]

Hmmm... the maths exam was an interesting experience.
Because at 1.20pm the fire alarm/sirens went off at my school. Very, very loudly. Then half the school went walking past the exam's classroom talking to go to the lawn. Then the examiner thought everyone should close their exam booklets, get up and go to the lawn (but not talk at all). However, another teacher came in and said we should sit. And so everyone started talking (but we weren't allowed to talk about the exam) and others left the room to go to the toilet. The sirens were still going for another 5 minutes, then the siren sound changed... and then it was be broken up with a soothing male voice stating 'evacuate as directed. Evacuate as directed'.


Very, very exciting. Finally the sirens stopped at around 1.30pm, and we were allowed another 15mins writing time because of it (we went till 2.15pm).

[/0]

Haha that sounds like fun

[monicak]

If I use the 2006 data in Mao's SS calculator [since the 06 exams were considered easy], will that give me an accurate indication of what SS I should get this year?

[Synesthetic]

also, could anyone explain 1 a iii  in a more algebrac way?? i got the answer same as ii though.....

Mean = integral over [2,6] of xf(x)dx, enter into calculator => 4.1333 hours

is that question 1 a iii?  i think it's the last question of 1......  i got that though...

Sorry:

What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Combinatorics solution:

• Given exactly 6 successes == given exactly 2 failures
• Our probability space has these 2 failures anywhere in the 8 attempts, so 8 choices for the first fail * 7 choices for the second fail = 56
• The event we want has these 2 failures anywhere in the last 4 attempts, so 4 choices for the first fail * 3 choices for the second fail = 12
• 
  

Yes, I saw that, i just don't understand...  it seems to be explaining a lot..

She shoots for goal 8 times;

Split the problem into two scenarios: the first four shots at goal[1], then the next four[2].

[1] Probability she scores each of the first four times
n=4, r=4, p=0.8 => Pr(X=4) = (4C4)(0.^4(0.2)^0 = 0.8^4 = 0.4096

[2] Probability she scores two of the next four times
n=4, r=2, p=0.8 => Pr(X=2) = (4C2)(0.^2(0.2)^2 = 0.1536

[1] x [2] = 0.063 to 3 d.p.

From part a)ii) of that question, Pr(X=6) = 0.2936

Conditional probability: 0.063 / 0.2936 = 0.214 to 3 d.p.

[/0]

Hmmm.... in the CAS exam I think there was only a) ii), not a) iii). Instead we had to do some markov chain stuff

[danieltennis]

Hmmm.... in the CAS exam I think there was only a) ii), not a) iii). Instead we had to do some markov chain stuff

CAS must be different then.