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HSC 2015 MX2 Permutations & Combinations Marathon (archive) (1 Viewer)

Thread starter Drsoccerball
Start date Aug 8, 2015

Status: Not open for further replies.

I

InteGrand

Well-Known Member

#381

Re: 2015 permutation X2 marathon

Not much point as far as I can see, but it was raised by someone and it was asked if the probability would remain the same.

seanieg89

Well-Known Member

#382

Re: 2015 permutation X2 marathon

$\mathcal{P}=\frac{1}{2ab}\int_{-a}^a \min(\frac{p^2}{4},b)\, dp$

seanieg89

Well-Known Member

#383

Re: 2015 permutation X2 marathon

I assume braintic wants the answer in case form, so I will let someone else do that

.

I

InteGrand

Well-Known Member

#384

Re: 2015 permutation X2 marathon

$Was the reason we could use ratio of areas for this that $P$ and $Q$ were uniformly distributed? In general, if they had arbitrary probability distributions we wouldn't be able to use simple ratio of areas (without transforming to something that has a uniform distribution or something), would we? Think we'd use a double integral.$

Last edited: Oct 7, 2015

B

braintic

Well-Known Member

#385

Re: 2015 permutation X2 marathon

InteGrand said:
$Was the reason we could use ratio of areas for this that $P$ and $Q$ were uniformly distributed? In general, if they had arbitrary probability distributions we wouldn't be able to use simple ratio of areas (without transforming to something that has a uniform distribution or something), would we? Think we'd use a double integral.$

If the distribution wasn't even in one variable, you would have to integrate the distribution itself. If it was uneven in both variables, you would need a double integral.
That is my understanding anyway - correct me if I'm wrong.

seanieg89

Well-Known Member

#386

Re: 2015 permutation X2 marathon

Basically, but areas is a restrictive way to think about it imo its is more of a ratio of sets in your probability space (space of outcomes).

In this case your probability space is a product measure space so in general we would need a double integral (weighted by the pdf or more generally the measure defining our choice of variables).

The product measure of the two uniform measures on the line though is the uniform measure on the plane. (uniform=Lebesgue) So measuring the set of desirable outcomes in this problem is just finding it's area as a ratio of the full rectangle of possible outcomes.

Paradoxica

-insert title here-

#387

Re: 2015 permutation X2 marathon

seanieg89 said:
Basically, but areas is a restrictive way to think about it imo its is more of a ratio of sets in your probability space (space of outcomes).

In this case your probability space is a product measure space so in general we would need a double integral (weighted by the pdf or more generally the measure defining our choice of variables).

The product measure of the two uniform measures on the line though is the uniform measure on the plane. (uniform=Lebesgue) So measuring the set of desirable outcomes in this problem is just finding it's area as a ratio of the full rectangle of possible outcomes.

Ok, so why was the question classified by OP as 2U? Double Integrals are beyond High School.

seanieg89

Well-Known Member

#388

Re: 2015 permutation X2 marathon

Paradoxica said:
Ok, so why was the question classified by OP as 2U? Double Integrals are beyond High School.

Note the words "in general", I was just giving a high brow overview of how this stuff generalises.

You do not need double integration for this particular question, as my single integral expression a few posts above shows.

Idk about 2U, but such examples of continuous probability exist in the standard textbooks. I believe one of the cambridge 3U books has an extension problem on the Buffon needle problem that works the same way.

Carrotsticks

Retired

#389

Re: 2015 permutation X2 marathon

seanieg89 said:
Note the words "in general", I was just giving a high brow overview of how this stuff generalises.

You do not need double integration for this particular question, as my single integral expression a few posts above shows.

Idk about 2U, but such examples of continuous probability exist in the standard textbooks. I believe one of the cambridge 3U books has an extension problem on the Buffon needle problem that works the same way.

It was touched on in the 2005 Maths Advanced paper Q10 (b).

D

Drsoccerball

Well-Known Member

#390

Re: 2015 permutation X2 marathon

Oh yes ! I remember this question because i couldn't do it...

R

rand_althor

Active Member

#391

Re: 2015 permutation X2 marathon

rand_althor said:
Question I saw: In a chess tournament, n women and 2n men participated, and each one of them played only one game with everybody else. The ratio of the number of games won by women to the number of games won by men is 7:5. Find the number of men that participated in the tournament if no game was over in a tie.
Answer: 6 men

The solution provided with the question:

$$Let $x$ be the number of matches in which women defeated men. The number of matches between two women is $\frac{n(n-1)}{2}$, the number of matches between two men is $\frac{2n(2n-1)}{2}$, and the number of matches between a man and a women is $2n^2$. The number of matches won by women is $\frac{n(n-1)}{2}+x$, and the number of matches won by men is $\frac{2n(2n-1)}{2}+2n^2-x$. Hence, $\left (\frac{n(n-1)}{2}+x \right ):\left (\frac{2n(2n-1)}{2}+2n^2-x \right )=7:5$. The last equation is equivalent to $8x=17n^2-3n$. Since $x \leq 2n^2$, we have $8x<16n^2$. This implies $17n^2-3n<16n^2$, i.e. $n(n-3)<0$. Hence the possible values of $n$ are 1,2 or 3. For $n=1$ and $n=2$ the equation $8x=17n-3n^2$ does not have integer solutions, and for $n=3$ we have $x=18$. Therefore six men participated in the tournament.$

Last edited: Oct 7, 2015

I

InteGrand

Well-Known Member

#392

Re: 2015 permutation X2 marathon

[The famous Monty Hall Problem]

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door which he knows has a goat, say No. 3. He then says to you, "Do you want to pick door No. 2?" What is the chance of winning the car if switch your choice?

B

braintic

Well-Known Member

#393

Re: 2015 permutation X2 marathon

InteGrand said:
[The famous Monty Hall Problem]

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door which he knows has a goat, say No. 3. He then says to you, "Do you want to pick door No. 2?" What is the chance of winning the car if switch your choice?

AND ..... you KNOW the host's strategy of only ever revealing a goat.

Paradoxica

-insert title here-

#394

Re: 2015 permutation X2 marathon

Monty Hall Problem with a twist: The remaining two doors are in quantum superpositions of both states at once. True 50:50 achieved.

D

Drsoccerball

Well-Known Member

#395

Re: 2015 permutation X2 marathon

braintic said:
AND ..... you KNOW the host's strategy of only ever revealing a goat.

So if he reveals thats its a goat then theres an even higher chance that if you switch you get the car right? Higher than 50% ?

I

InteGrand

Well-Known Member

#396

Re: 2015 permutation X2 marathon

Drsoccerball said:
So if he reveals thats its a goat then theres an even higher chance that if you switch you get the car right? Higher than 50% ?

Yeah

D

Drsoccerball

Well-Known Member

#397

Re: 2015 permutation X2 marathon

InteGrand said:
Yeah

2/3?

I

InteGrand

Well-Known Member

#398

Re: 2015 permutation X2 marathon

Drsoccerball said:
2/3?

Yep

D

Drsoccerball

Well-Known Member

#399

Re: 2015 permutation X2 marathon

InteGrand said:
Yep

ahahahah thats awesome... I never knew probability works like that if someone knows the results..

I

InteGrand

Well-Known Member

#400

Re: 2015 permutation X2 marathon

Drsoccerball said:
ahahahah thats awesome... I never knew probability works like that if someone knows the results..

Haha. Probability is all about the knowledge of the person.

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