HSC 2016 MX2 Combinatorics Marathon (archive) (1 Viewer)

InteGrand · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

I'd interpreted it as that too, main reason being that braintic had referred to a "length" in part (a), so I thought he was talking about this length again in part (b).

Paradoxica · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

braintic said:
It hadn't occurred to me that my question was ambiguous. My intention was that the original interval remains length 1, and the length of the middle segment would change to get a probability of 1/3.

That's alright, we got both answers anyway!

And we can take this as a point to make for future questions: Be wary of ambiguity.

InteGrand · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

braintic said:
Open to all. And the method IS related to a past HSC question.

Is this past HSC Q. the one about waiting times (I think might have been the last Q. of a 2U paper)?

braintic · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

InteGrand said:
Is this past HSC Q. the one about waiting times (I think might have been the last Q. of a 2U paper)?

Yep, that's the one.

Paradoxica · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

A and B are random, independent, uniformly distributed points on the interval [0,1], and divide the interval into three segments.

What is the probability that the three segments can form a triangle?

braintic · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Just checking - is the answer 1/4 ?
(If it is, I'll leave it for someone else to provide a solution)

Paradoxica · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

braintic said:
Just checking - is the answer 1/4 ?
(If it is, I'll leave it for someone else to provide a solution)

Wow, leaving things as exercises to readers.

You'd make a great stereotypical publisher.

braintic · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Paradoxica said:
Wow, leaving things as exercises to readers.

You'd make a great stereotypical publisher.

Well I am the teacher here. I have to limit my showing off for the benefit of the students .... difficult as that might be.

Paradoxica · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

braintic said:
Well I am the teacher here. I have to limit my showing off for the benefit of the students .... difficult as that might be.

just noticed your post count.

one more for this year XD

braintic · Apr 25, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Paradoxica said:
just noticed your post count.

one more for this year XD

Hmm ... not sure how to interpret that comment.

Ambility · Apr 30, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

$$Find the number of ways to arrange 4 letters selected from the word DIVERSITY.$

DatAtarLyfe · Apr 30, 2016

HSC 2016 MX2 Combinatorics Marathon

Ambility said:
$$Find the number of ways to arrange 4 letters selected from the word DIVERSITY.$

What's the final answer? I think my answer may be too simple

Ambility · Apr 30, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

DatAtarLyfe said:
What's the final answer? I think my answer may be too simple

Haha I'm actually putting this up because I wasn't sure about my answer either. I got 1932 ways, what did you get?

Drsoccerball · Apr 30, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Ambility said:
Haha I'm actually putting this up because I wasn't sure about my answer either. I got 1932 ways, what did you get?

Yes, your answer's correct.

InteGrand · May 1, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

$\noindent An urn contains $r$ red marbles and $g$ green marbles, where $r,g$ are fixed positive integers. Alice randomly draws $n$ marbles from the urn without replacement (where $n$ is a fixed positive integer, at most $r+g$). For $j\in \left \{0,1,2,\ldots , n\}$, let $G_j$ be the event that the $j$th draw is green. Find $\mathbb{P}\left(G_{j}\right)$ for each $j$. (This is a \emph{prior probability}, before we have begun drawing or noting any draws.)$

$\noindent \textit{Bonus.} Let $X$ be the number of greens drawn from the $n$ draws. Find an expression for $\mathbb{P}\left(X=k\right)$ for each integer $k$ from $\max \left(0, n -r\right)$ to $\min \left(n, g\right)$ inclusive, and explain the bounds here for $k$.$

Paradoxica · May 1, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

$$\noindent Drsoccerball awakens in the centre of a perfectly square paddock out in the countryside.$\\$The side length of the paddock fence is exactly 2 kilometers.$\\$Confused, he picks a random direction and begins walking in a straight line until he reaches the fence, at which point, he is either instantaneously and magically whisked away by a tornado into oblivion, or watches as the tornado vanishes before his very eyes.$\\$a) Suppose that any point along the fence less than or exactly $r$ metres away, where $1000 \leq r \leq 1000\sqrt{2}$, from the center is completely safe from the tornado. What, then, is the probability that Drsoccerball survives?$\\$b) What is the expected distance he travels, given that he picks an angle which lets him survive the tornado?$\\$

KingOfActing · May 1, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Paradoxica said:
$$\noindent Drsoccerball awakens in the centre of a perfectly square paddock out in the countryside.$\\$The side length of the paddock fence is exactly 2 kilometers.$\\$Confused, he picks a random direction and begins walking in a straight line until he reaches the fence, at which point, he is either instantaneously and magically whisked away by a tornado into oblivion, or watches as the tornado vanishes before his very eyes.$\\$a) Suppose that any point along the fence less than or exactly $r$ metres away, where $1000 \leq r \leq 1000\sqrt{2}$, from the center is completely safe from the tornado. What, then, is the probability that Drsoccerball survives?$\\$b) What is the expected distance he travels, given that he survives the tornado?$\\$

Part a: (explanation to come as soon as I type it)

$P(\text{He survives}) = \frac{2\cos^{-1}\left(\frac{2000000 - r^2}{r^2}\right)}{\pi}$

KingOfActing · May 1, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

$$Consider the attached diagram. Our probability (assuming uniform distribution for the random direction choosen) is $\frac{4\theta}{2\pi} = \frac{2\theta}{\pi}\\\\$Using Pythagoras's theorem (twice), we get:$ \left(1 - \frac{x}{2}\right)^2 + \left(1+\frac{x}{2} \right )^2 = 2r^2 \iff x^2 = 4r^2-4\\$Now, using the cosine rule:$\\x^2 = 2r^2 - 2r^2\cos\theta \iff\theta = \cos^{-1}\left(\frac{2-r^2}{r^2}\right)\\$However, we must match the units, so $ r \mapsto \frac{r}{1000}\\$Putting all these results together, we get:\\$P(\text{Survival}) = \frac{2\cos^{-1}\left(\frac{2000000-r^2}{r^2}\right)}{\pi}$

KingOfActing · May 2, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

For part b I'm not sure if I did it right, if the result is right I'll post the working:

$E[r] = \frac{4000\ln(1+\sqrt 2)}{\pi}$

KingOfActing · May 2, 2016

Re: HSC 2016 MX2 Combinatorics Marathon

Working:

$$The function we calculated in part a is actually the CDF, so the expected value is:$ \int_{1000}^{1000\sqrt{2}}r \frac{d}{dr}\left[\frac{2\cos^{-1}\left(\frac{2000000-r^2}{r^2}\right)}{\pi}\right]\,dr = \int_{1000}^{1000\sqrt{2}}\frac{4000}{\pi\sqrt{r^2-1000000}}\,dr=\frac{4000\ln(1+\sqrt2)}{\pi}$

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