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HSC 2013 MX2 Marathon (archive) (7 Viewers)

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harryharper

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Re: HSC 2013 4U Marathon

Part i: first step change it to 11/8 + 0.5*Sum from r=3 to infinite of (2r-1)/(r^3-r). This comes from 'changing the perspective' from summing sums of 1/k terms to sums of 1/((k+2)k) terms.
Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4 as the answer.
 
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Sy123

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Re: HSC 2013 4U Marathon

Part i: first step change it to 11/8 + Sum from r=3 to infinite of (2r-1)/(r^3-r)
Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4 as the answer.
How did you get (2r-1)/(r^3-r), that is the correct answer though, I split the 1/k(k+2) to 1/2 ( 1/k - 1/(k+2)), then manipulated and grouped certain terms to telescope twice.
 

harryharper

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Re: HSC 2013 4U Marathon

Intermediate steps included:

11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....)

Which equals

11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r)

Sorry no Latex - still working on my skills in that department.
 

Sy123

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Re: HSC 2013 4U Marathon

Intermediate steps included:

11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....)

Which equals

11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r)

Sorry no Latex - still working on my skills in that department.

Ah yep ok that is quite similar to what I did.
 

harryharper

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Re: HSC 2013 4U Marathon

This one's the triangle inequality right? Just need a little bit of induction to extend it.
 

Sy123

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Re: HSC 2013 4U Marathon

Ohhh I see what you mean now, nice work.

I squared both sides and cancelled out the a_k^2 and b_k^2 terms, and then by grouping the terms

proving



Which is done by squaring both sides and expanding, to yield



Which is just the AM-GM inequality.

Your method is more elegant though lol

----



 

harryharper

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Re: HSC 2013 4U Marathon



No squares between consecutive squares.
 
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harryharper

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Re: HSC 2013 4U Marathon

There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any.

i) What is the probability of me drawing out K red marbles?

ii) What is the most likely number of red marbles I will draw out?

iii) What is the average number of red marbles I will draw out?
 
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Sy123

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Re: HSC 2013 4U Marathon

There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any.

i) What is the probability of me drawing out K red marbles?

ii) What is the most likely number of red marbles I will draw out?

iii) What is the average number of red marbles I will draw out?
Here is an attempt:













 
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harryharper

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Re: HSC 2013 4U Marathon

Here is an attempt:













i) is correct.

ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1
 

Sy123

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Re: HSC 2013 4U Marathon

i) is correct.

ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1
Ahh ok, I'm not too sure about how to interpret 'average' though..

If I did:



And took the average probability:



And then, if:

(so its bounded by two consecutive probabilities)

Then the one its closest to, (say its closest to P(m)), that means m is the average number of balls????
 

harryharper

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Re: HSC 2013 4U Marathon

Ahh ok, I'm not too sure about how to interpret 'average' though..

If I did:



And took the average probability:



And then, if:

(so its bounded by two consecutive probabilities)

Then the one its closest to, (say its closest to P(m)), that means m is the average number of balls????










 
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Sy123

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Re: HSC 2013 4U Marathon

Oh ok

As for calculation of E(x)





Consider the identity:



Equate co-efficient of on both sides, where for the first binomial, take the differentiated version of as its expansion, we then yield the sum:



Hence the expected probability is:

 

harryharper

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Re: HSC 2013 4U Marathon

Yep - just missing the 'r' from the RHS of your identity in the last two lines. Final answer should be nr/m.
 
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