HSC 2016 Maths Marathon (archive) (1 Viewer)

[Green Yoda]

Re: HSC 2016 2U Marathon

Nope, it's not a symmetry probability as Sky has the advantage of going first.

both wrong?

 

[InteGrand]

Re: HSC 2016 2U Marathon

a) 1/36
b) bit confused here as the game will end as proposed in the question so my answer of sky winning is 1/2 as there are 2 players. (might be wrong).

 

[leehuan]

Re: HSC 2016 2U Marathon

Basically what InteGrand said. Also your part (a) was fine.
____________________________________
NEXT QUESTION:
Troll time

 

[davidgoes4wce]

Re: HSC 2016 2U Marathon

its k, join the club.
Next question, for 2016ers ONLY (wanna see how many can get it, found it relatively easy)
View attachment 32641

Would I be classified as a non-2016er now?

 

[Sien]

Re: HSC 2016 2U Marathon

Would I be classified as a non-2016er now?

You're not doing the hsc 2016, therefore you're not a 2016er lol

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[Green Yoda]

Re: HSC 2016 2U Marathon

You're not doing the hsc 2016, therefore you're not a 2016er lol

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I'm not a 2016er..can i still answer?

 

[Speed6]

Re: HSC 2016 2U Marathon

I'm not a 2016er..can i still answer?

Yes, you can answer Rathin

 

[qwert73]

Re: HSC 2016 2U Marathon

A school softball team has a probability of 0.2 of winning a match.
i) Find the probability the team wins exactly one of its first 2 matches
ii) What is the least number of consecutive matches the team must play to be 90% certain that it will win at least one match

hint(you need sum of infinity)

 

[davidgoes4wce]

Re: HSC 2016 2U Marathon

My thinking for question

i) 2C1 x (0.20)^1 x (0.80)^1 = 0.32

ii) I drew a tree diagram for this question and equated the probability of winning at least 0.90 , after losing in 'n' attempts
0.90= 1-(0.80)^n

n=10.32

A minimum number of times to ensure that he has at least a 90% chance of winning, will be 11 times.

 

[davidgoes4wce]

Re: HSC 2016 2U Marathon

I also thought about if you rolled out 2 attempts

your sample space would be :{WW, WL, LW, LL}

The probability of LL= 0.80 x 0.80= 0.64

Your chances of 'winning at least once' after 2 attempts is 0.36

You want to get that probability of losing to be under 0.10, which implies that the probability of winning is at least once is 0.90 after 11 attempts.

 

[Sien]

Re: HSC 2016 2U Marathon

My thinking for question

i) 2C1 x (0.20)^1 x (0.80)^1 = 0.32

ii) I drew a tree diagram for this question and equated the probability of winning at least 0.90 , after losing in 'n' attempts
0.90= 1-(0.80)^n

n=10.32

A minimum number of times to ensure that he has at least a 90% chance of winning, will be 11 times.

You don't do combinations in 2u lol

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[davidgoes4wce]

Re: HSC 2016 2U Marathon

You don't do combinations in 2u lol

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OK, then you could use a tree diagram then?

P(WL)+P(LW)=0.20x0.80+0.80x0.20=0.32

Still get the same value anyway

 

[Sien]

Re: HSC 2016 2U Marathon

OK, then you could use a tree diagram then?

P(WL)+P(LW)=0.20x0.80+0.80x0.20=0.32

Still get the same value anyway

Yea I'm just saying

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[leehuan]

Re: HSC 2016 2U Marathon

Yes, in 2U you use the tree diagram. Avoid using combinatorics in a 2U marathon where possible; even a tree diagram will suffice

Nobody's going to answer this troll question? It's honestly REALLY easy.

 

[InteGrand]

Re: HSC 2016 2U Marathon

Yes, in 2U you use the tree diagram. Avoid using combinatorics in a 2U marathon where possible; even a tree diagram will suffice



Nobody's going to answer this troll question? It's honestly REALLY easy.

 

[leehuan]

Re: HSC 2016 2U Marathon

Lol.
-----------------------------------------------
NEXT QUESTION
Exercise:

 

[Green Yoda]

Re: HSC 2016 2U Marathon

Lol.
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NEXT QUESTION
Exercise:

Using power rule.
f(x) 1/x = x^-1. So f'(x) = -x^-2. Therefore f'(x) = -1/x^2

 

[InteGrand]

Re: HSC 2016 2U Marathon

Using power rule.
f(x) 1/x = x^-1. So f'(x) = -x^-2. Therefore f'(x) = -1/x^2

This is not first principles.

 

[Green Yoda]

Re: HSC 2016 2U Marathon

This is not first principles.

my bad.

y ' = lim (h → 0) (f(x+h) - f(x)) / h
= lim (h → 0) (1/(x+h) - 1/x) / h
= lim (h → 0) (x/x(x+h) - (x + h)/x(x + h)) / h
= lim (h → 0) (-h / x(x + h)) / h
= lim (h → 0) -1 / x(x + h)
= -1/ x²

 

[RachelGreen]

Re: HSC 2016 2U Marathon