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HSC 2015 MX1 Marathon (archive) (1 Viewer)

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rand_althor

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Re: HSC 2015 3U Marathon

From a box with red and blue balls we randomly choose two balls. Assume that the box has at least two balls and at least one of them is blue. The probability that the chosen balls are both red is five times the probability that the chosen balls are both blue and the probability that the chosen balls are of different colour is six times the probability that the chosen balls are both blue. How many red balls are in the box?
 

rand_althor

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Re: HSC 2015 3U Marathon

A mathematics contest consists of four problems. Each of the six member team from Central High School is assigned to work on exactly one of the four problems. If each of the four problems is worked on by at least one of the team, in how many different ways can the assignment of team members to problems be accomplished?
 

sida1049

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Re: HSC 2015 3U Marathon

A mathematics contest consists of four problems. Each of the six member team from Central High School is assigned to work on exactly one of the four problems. If each of the four problems is worked on by at least one of the team, in how many different ways can the assignment of team members to problems be accomplished?
Number of ways each problem is assigned to exactly one member: 6P4 = 360

Number of ways the last two members of the team can be assigned a problem each: 6x6

Total = 36 X 360 = 12960

(Messed up the first time... hahah combinatorics is my weakest point. Not particularly confident about this answer either.)
 

InteGrand

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Re: HSC 2015 3U Marathon

Number of ways each problem is assigned to exactly one member: 6P4 = 360

Number of ways the last two members of the team can be assigned a problem each: 6x6

Total = 36 X 360 = 12960

(Messed up the first time... hahah combinatorics is my weakest point. Not particularly confident about this answer either.)
The answer can't be 12960, because this is greater than the total number of ways to assign exactly one question to 6 people without restrictions, (i.e. without ensuring that every question has at least one person working on it), which is 46 = 4096.
 

sida1049

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Re: HSC 2015 3U Marathon

The questions can be assigned either as [1,1,1,3] or [1,1,2,2]:

[1,1,1,3]: 4 * (6C3 * 3 * 2 * 1) = 480
[1,1,2,2]: 4C2 * (6C2 * 4C2 * 2) = 1080

Therefore, total ways = 1560
Ahh thank you. I can now go to sleep. For me at least, the small topic of combinatorics in Extension 1 is the hardest part across the entire HSC mathematics courses... it's gonna take up all my post-HSC break.
 

Drsoccerball

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Re: HSC 2015 3U Marathon

The questions can be assigned either as [1,1,1,3] or [1,1,2,2]:

[1,1,1,3]: 4 * (6C3 * 3 * 2 * 1) = 480
[1,1,2,2]: 4C2 * (6C2 * 4C2 * 2) = 1080

Therefore, total ways = 1560
Zlatman for president.
 

rand_althor

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Re: HSC 2015 3U Marathon

Here's another probability one:
Two points are picked at random on the unit circle . What is the probability that the chord joining the two points has length at least 1?

Idk if it is 3U difficulty or not. If it isn't I'll post it to the X2 Perms/Combs Marathon.
 

Drsoccerball

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Re: HSC 2015 3U Marathon

Here's another probability one:
Two points are picked at random on the unit circle . What is the probability that the chord joining the two points has length at least 1?

Idk if it is 3U difficulty or not. If it isn't I'll post it to the X2 Perms/Combs Marathon.
I think this is similar to Braintic's question. Since there was a 2U question on it im guessing it can be 3U difficulty.
 

InteGrand

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Re: HSC 2015 3U Marathon

Here's another probability one:
Two points are picked at random on the unit circle . What is the probability that the chord joining the two points has length at least 1?

Idk if it is 3U difficulty or not. If it isn't I'll post it to the X2 Perms/Combs Marathon.
If I recall braintic's question correctly, it was exactly equivalent to this (otherwise a variant, but same idea). The answer varies depending on the method of randomly selecting the two points. Since the method is unspecified, there is no definitive answer to the question as posed.
 

rand_althor

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Re: HSC 2015 3U Marathon

If I recall braintic's question correctly, it was exactly equivalent to this (otherwise a variant, but same idea). The answer varies depending on the method of randomly selecting the two points. Since the method is unspecified, there is no definitive answer to the question as posed.
What methods of randomly selecting the two points are there? I thought it would be enough to know both points lie on the circle.
 

Drsoccerball

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Re: HSC 2015 3U Marathon

Here's another probability one:
Two points are picked at random on the unit circle . What is the probability that the chord joining the two points has length at least 1?

Idk if it is 3U difficulty or not. If it isn't I'll post it to the X2 Perms/Combs Marathon.
I got 1/5 ...?
 

InteGrand

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Re: HSC 2015 3U Marathon

braintic's question was about picking a "random" chord, which is equivalent to picking a "random" pair of points. The method of randomly selecting them must be specified in order to obtain a definitive answer:

 

rand_althor

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Re: HSC 2015 3U Marathon

braintic's question was about picking a "random" chord, which is equivalent to picking a "random" pair of points. The method of randomly selecting them must be specified in order to obtain a definitive answer:
Ah okay I see what you mean. From the answer I'm pretty sure they intend for you to use the first method you've quoted.
 
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