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HSC 2013 MX2 Marathon (archive) (4 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon





Can u see any mistakes?
Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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HeroicPandas

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Re: HSC 2013 4U Marathon

Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)
YESSSSSSSS!!!!!! :D You have made my day Sy!
 

Sy123

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Re: HSC 2013 4U Marathon

Isn't it meant to be 2x not 2(x+1)?
Ah yes, I'm making too many mistakes at the moment :/
However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?
 

bleakarcher

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Re: HSC 2013 4U Marathon

Ah yes, I'm making too many mistakes at the moment :/
However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?
Yeah, both are correct expressions for the primitive.
 

HeroicPandas

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Re: HSC 2013 4U Marathon

Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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Question:

 
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Drongoski

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Re: HSC 2013 4U Marathon

Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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Just trying!












 
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seanieg89

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Re: HSC 2013 4U Marathon

A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?
 

anomalousdecay

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Re: HSC 2013 4U Marathon

A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?

a) P($20) = (18/37)^10
P($0) = (19/37)^10

b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.


P.S. The Probability of me getting this question wrong is 40% :p as I thought that it may be possible to use another method to work it out, but I thought this method is most correct.
Please correct me if I am wrong. We all learn from our mistakes.
 
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Sy123

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Re: HSC 2013 4U Marathon

a) P($20) = (18/37)^10
P($0) = (19/37)^10

b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.


P.S. The Probability of me getting this question wrong is 40% :p as I thought that it may be possible to use another method to work it out, but I thought this method is most correct.
Please correct me if I am wrong. We all learn from our mistakes.
But he can lose say 9$ then win 19 to win, your probability is for when he wins 10 times in a row, it is merely one case.
 

anomalousdecay

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Re: HSC 2013 4U Marathon

But he can lose say 9$ then win 19 to win, your probability is for when he wins 10 times in a row, it is merely one case.
Yeah that makes more sense.
So then would it be......

a)
P($20) = 10C0 (18/37)^10(19/37) + 10C1(18/37)^11(19/37) + 10C2(18/37)^12(19/37)^2 + 10C3(18/37)^13(19/37)^3 +10C4(18/37)^14(19/37)^4 + 10C5(18/37)^15(19/37)^5 + 10C6(18/37)^16(19/37)^6 +10C7(18/37)^17(19/37)^7 + 10C8(18/37)^18(19/37)^8 + 10C9(18/37)^19(19/37)^9

P($0) = 1- P($20)


b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.

?????? For some reason I am thinking that I missed something again :confused2:.
 
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