VCE Methods Question Thread!

[ahat]

Hi, can someone explain to me how to work out the value of a combinatoric by hand?

I know the formula is n!/r!(n-r)! but I'm not sure how to work out the values. Thanks!

let's say n = 5 and r = 3

n! = 5 x 4 x 3 x 2 x 1
r! = 3 x 2 x 1
(n - r)! = 2! = 2 x 1

therefore,
(5 x 4 x 3 x 2 x 1)/((3 x 2 x 1)(2 x 1))
Before we even start multiplying, simplify expression

(5 x 4)/(2 x 1) = 10

[sasa]

Oh my gosh! Thank you. Would you need to expand though? Couldn't you just leave it in this case as 2/(2pi+2)^2?

[Henreezy]

Oh my gosh! Thank you. Would you need to expand though? Couldn't you just leave it in this case as 2/(2pi+2)^2?

Yes, you could leave it in that form.
I don't know why VCAA didn't show the alternative answer in the assessor's report.
Expansion is just to show you that both answers are the same, just that one answer is in simplest form.

[Henreezy]

Okay I have a question of my own, it's bugged me for a while but I don't get it at all (wow have I left it late).

"Find a such that Pr(Z>a) = Pr(X<5)"
Given that the mean = 10 and X = 5, I did
Z =(5-10)/2, giving me Z= -5/2

Why is a = positive 5/2?
Is it because Pr(Z>a) is the same magnitude as Pr(X<5), but the negative Z value is giving me negative a?
If it was Pr(Z>-a), it wouldn't be the right probability, it would be a crap load more than 5 I'm assuming (from visualising the distribution curve).

[Damoz.G]

Okay I have a question of my own, it's bugged me for a while but I don't get it at all (wow have I left it late).

"Find a such that Pr(Z>a) = Pr(X<5)"
Given that the mean = 10 and X = 5, I did
Z =(5-10)/2, giving me Z= -5/2

Why is a = positive 5/2?
Is it because Pr(Z>a) is the same magnitude as Pr(X<5), but the negative Z value is giving me negative a?
If it was Pr(Z>-a), it wouldn't be the right probability, it would be a crap load more than 5 I'm assuming (from visualising the distribution curve).

I think its positive because you work from left to right, and its symmetrical.

[Sanguinne]

how do i do question 2e in vcaa 2011 exam 2

[Henreezy]

I think its positive because you work from left to right, and its symmetrical.

Ah yeah, plus it wouldn't make a great deal of sense if it was Pr(Z>-5/2) as Pr(X<5) would be a much smaller quantity (less than 0.5). I get the whole symmetry deal, it's just these specific questions always trip me up. At the very least I would have got one mark out of two .
Thx Damoz!

[Conic]

how do i do question 2e in vcaa 2011 exam 2

It states a random chocolate is selected from the bin. It is found to have taken longer than 3 seconds to produce. The chance of it coming from A or B is equal, since there are equal numbers and the selection is "random".  Since it has been found to have taken longer than 3 seconds, there is a condition. This condition means we are looking for (T is the time taken).

First, we have to find , which is . Using CAS this is .

Now we need to find . The time can take longer than 3 seconds with both machines , so:



We worked out and , so now we can find :

[Damoz.G]

Ah yeah, plus it wouldn't make a great deal of sense if it was Pr(Z>-5/2) as Pr(X<5) would be a much smaller quantity (less than 0.5). I get the whole symmetry deal, it's just these specific questions always trip me up. At the very least I would have got one mark out of two .
Thx Damoz!

No worries.

I used to never understand why when we covered the topic, and just flipped it because I assumed we had to. But now, after doing lots of Trials, I understand why, which is exactly the same thing as what you explained.

[Sanguinne]

It states a random chocolate is selected from the bin. It is found to have taken longer than 3 seconds to produce. The chance of it coming from A or B is equal, since there are equal numbers and the selection is "random".  Since it has been found to have taken longer than 3 seconds, there is a condition. This condition means we are looking for (T is the time taken).

First, we have to find , which is . Using CAS this is .

Now we need to find . The time can take longer than 3 seconds with both machines , so:



We worked out and , so now we can find :

thanks conic, very detailed answer 

[Conic]

[Alwin]

What's the symbol for the determinant of a matrix?

Like a triangle is the symbol for the descriminant of a quadratic, what's the one for determinant?

Or

(note the straight lines, not ( ) or [ ] )

Which is less popular in VCE because co-expansion isn't taught

[Henreezy]

Not really a question but a handy hint for tomorrow!
If you've noticed, integration problems often have really 'sneaky' wording sometimes.
ONLY when it says find "an" anti-derivative, can you exclude +c; it's implied that it's a part of some family of functions so it's optional to put +c or not.
If it says integrate this function, you put it on. This primarily relates to 'hence find an anti-derivative for function', in which you use your method from part a to separate and find an integral for said function.

[Damoz.G]

(note the straight lines, not ( ) or [ ] )

Which is less popular in VCE because co-expansion isn't taught

That is news to me!!!

I used to always write it out as a matrix equation for a Determinant. Probably would lost marks if I did that on the Exams, wouldn't I? =/

[Damoz.G]

Not really a question but a handy hint for tomorrow!
If you've noticed, integration problems often have really 'sneaky' wording sometimes.
ONLY when it says find "an" anti-derivative, can you exclude +c; it's implied that it's a part of some family of functions so it's optional to put +c or not.
If it says integrate this function, you put it on. This primarily relates to 'hence find an anti-derivative for function', in which you use your method from part a to separate and find an integral for said function.

I've noticed that. I think I saw it in last year's Examiner's Report, the Chief Examiner said "the +c was not necessary".