Green Yoda
Hi Φ
- Joined
- Mar 28, 2015
- Messages
- 2,859
- Gender
- Male
- HSC
- 2017
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
HSC 2016 Maths Marathon (archive) (2 Viewers)
- Thread starter Speed6
- Start date
- Status
davidgoes4wce
Well-Known Member
Re: HSC 2016 2U Marathon
Sent from my D6503 using Tapatalk
You're not doing the hsc 2016, therefore you're not a 2016er lol
Sent from my D6503 using Tapatalk
Green Yoda
Hi Φ
- Joined
- Mar 28, 2015
- Messages
- 2,859
- Gender
- Male
- HSC
- 2017
Speed6
Retired '16
- Joined
- Jul 31, 2014
- Messages
- 2,949
- Gender
- Male
- HSC
- N/A
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)
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
Well-Known Member
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.
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
Well-Known Member
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.
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.
Re: HSC 2016 2U Marathon
Sent from my D6503 using Tapatalk
You don't do combinations in 2u lol
Sent from my D6503 using Tapatalk
davidgoes4wce
Well-Known Member
Re: HSC 2016 2U Marathon
P(WL)+P(LW)=0.20x0.80+0.80x0.20=0.32
Still get the same value anyway
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
leehuan
Well-Known Member
- Joined
- May 31, 2014
- Messages
- 5,805
- Gender
- Male
- HSC
- 2015
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
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. \left( 1-\cos ^{ 2 }{ x } \right) \left( 1+\tan ^{ 2 }{ x } \right) \left( 1+\cot ^{ 2 }{ x } \right) dx } )
Green Yoda
Hi Φ
- Joined
- Mar 28, 2015
- Messages
- 2,859
- Gender
- Male
- HSC
- 2017
Re: HSC 2016 2U Marathon
f(x) 1/x = x^-1. So f'(x) = -x^-2. Therefore f'(x) = -1/x^2
Using power rule.Lol.
-----------------------------------------------
NEXT QUESTION
Exercise:
 to show that $\\ \frac { d }{ dx } \left( \frac { 1 }{ x } \right) =-\frac { 1 }{ { x }^{ 2 } } )
f(x) 1/x = x^-1. So f'(x) = -x^-2. Therefore f'(x) = -1/x^2
Green Yoda
Hi Φ
- Joined
- Mar 28, 2015
- Messages
- 2,859
- Gender
- Male
- HSC
- 2017
Re: HSC 2016 2U Marathon
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
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
Last edited:
RachelGreen
Member
- Joined
- Oct 13, 2015
- Messages
- 84
- Gender
- Male
- HSC
- 2017
Re: HSC 2016 2U Marathon
Last edited:
- Status