UNSW competition senior division (1 Viewer)

[Sy123]

http://www.maths.unsw.edu.au/sites/default/files/final_senior_51.pdf

Problem 1, Scenario 1.

What I was planning to do was, to recursively find the probability of finding people at E on the first night.

So,









Therefore we arrive at the recurrence relation:



However, I don't know how to find the limiting value of this sequence, making: doesn't work in this case....

Any suggestions?

Also, if it comes to the point where evaluating P(1) -> P(5) is necessary, we can just take the cases of if he picks A first, B, C, D, or E first to evaluate the first 5 constants as necessary....

Thanks

 

[iJimmy]

Holy fk is this what we are expected to do at Uni GOML!!!!!!!

 

[Makematics]

Holy fk is this what we are expected to do at Uni GOML!!!!!!!

that's from a high school maths comp lolz

 

[Sy123]

Also can anyone verify my answers to Problem 4:





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Also does anyone have copies of the previous UNSW maths competition papers (preferably with solutions), it says its the 51st competition and yet they only put 1 on their site :/

 

[seanieg89]

http://www.maths.unsw.edu.au/sites/default/files/final_senior_51.pdf

Problem 1, Scenario 1.

What I was planning to do was, to recursively find the probability of finding people at E on the first night.

So,









Therefore we arrive at the recurrence relation:



However, I don't know how to find the limiting value of this sequence, making: doesn't work in this case....

Any suggestions?

Also, if it comes to the point where evaluating P(1) -> P(5) is necessary, we can just take the cases of if he picks A first, B, C, D, or E first to evaluate the first 5 constants as necessary....

Thanks

Judging by the wording of the question, I think you are allowed to assume that there IS a long-term probability associated to each town. ie lim P(person stays in town X on n-th night) exists for X=A,B,C,D,E. (Even if you just assume it for X=E, this gives you convergence for the other towns.)

So just let a,b,c,d,e be the long term probabilities and formulate the transition conditions as a system of linear equations.

 

[Sy123]

Judging by the wording of the question, I think you are allowed to assume that there IS a long-term probability associated to each town. ie lim P(person stays in town X on n-th night) exists for X=A,B,C,D,E. (Even if you just assume it for X=E, this gives you convergence for the other towns.)

So just let a,b,c,d,e be the long term probabilities and formulate the transition conditions as a system of linear equations.

I get the first part, but what do you mean by what is bolded?

 

[seanieg89]

I get the first part, but what do you mean by what is bolded?

He stays in town B on the n-th night iff he stays in town A on the (n-1)-th night, so p_n(A)=p_{n-1}(B). Letting n-> infty gives a = b.

Do something like this for a,b,c,d,e, you will get a linear equation for each one. Just solve these.

Also, I think your answers to 4 are incorrect, I got 8/9 and 65/72.

 

[obliviousninja]

You know its gg when Sy asks for suggestions.

 

[RealiseNothing]

I'm pretty sure when I did this question, I found the probability of him being in each town as a "weighting" or something.

 

[seanieg89]

I'm pretty sure when I did this question, I found the probability of him being in each town as a "weighting" or something.

What do you mean?

 

[RealiseNothing]

What do you mean?

Something like etc

like I found the probability of him being in each town in terms of each other? It's been a while though.

 

[seanieg89]

Something like etc

like I found the probability of him being in each town in terms of each other? It's been a while though.

Ah okay, that is exactly what I suggested to Sy. It comes out v. fast.

 

[RealiseNothing]

I just thought of something, since there's a 50/50 chance of him going back to town D, then wouldn't it just cycle (long term) as so:

A > B > C > D > E > A > B > C > D > E > D > E

meaning the answer would be

 

[Sy123]

He stays in town B on the n-th night iff he stays in town A on the (n-1)-th night, so p_n(A)=p_{n-1}(B). Letting n-> infty gives a = b.

Do something like this for a,b,c,d,e, you will get a linear equation for each one. Just solve these.

Also, I think your answers to 4 are incorrent, I got 8/9 and 65/72.

Ohh I see, is this similar to what you had in mind:











Add them all side by side, take limit n to infinity

a+b+c+d+e = 1

1 = \frac{7}{2} L

????

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Also for problem 4, can you explain reasoning for part (a)?

From what I can see, the fact that the coin is tossed with 3 heads, is not dependent on the probability that he picked the biased coin?

Thanks for the help

 

[Sy123]

I just thought of something, since there's a 50/50 chance of him going back to town D, then wouldn't it just cycle (long term) as so:

A > B > C > D > E > A > B > C > D > E > D > E

meaning the answer would be

lol well that was another fail again on my part

I'm so bad at probability

 

[RealiseNothing]

From what I can see, the fact that the coin is tossed with 3 heads, is not dependent on the probability that he picked the biased coin?

Thanks for the help

Nah it does influence the probability. The answer isn't 1/2.

 

[RealiseNothing]

lol well that was another fail again on my part

I'm so bad at probability

I don't know if that method is correct though. I think you're answer of 2/7 might be right because I remember when I did it a while ago I got a fraction with 7 in the denominator.

 

[RealiseNothing]

Also, I think your answers to 4 are incorrect, I got 8/9 and 65/72.

I agree with this.

 

[Sy123]

Nah it does influence the probability. The answer isn't 1/2.

Ah alright, can you explain how to get seanieg's answer?

 

[RealiseNothing]

4)a) If you chose the 2 headed coin, the probability of flipping 3 heads is 1. If you chose the fair coin, the probability of flipping 3 heads is 1/8.

So since you did flip 3 heads, it is 8 times more likely that it is the 2 headed coin (1 vs 1/8). Hence the probabilities must be 8/9 and 1/9.