Green Yoda
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HSC 2016 Maths Marathon (archive) (5 Viewers)
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davidgoes4wce
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Re: HSC 2016 2U Marathon
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You're not doing the hsc 2016, therefore you're not a 2016er lol
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Green Yoda
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Speed6
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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
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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
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You don't do combinations in 2u lol
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davidgoes4wce
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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
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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
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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
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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
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Re: HSC 2016 2U Marathon
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