HSC 2015 MX1 Marathon (archive) (3 Viewers)

Paradoxica · Oct 7, 2015

Re: HSC 2015 3U Marathon

Doubt such a question is HSC style. Looks like Brilliant.org style to me, since it asks for m+n

rand_althor · Oct 7, 2015

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?

kawaiipotato · Oct 7, 2015

Re: HSC 2015 3U Marathon

$let x be the amount of red balls where x \geq 1$
$let y be the amount of blue balls where y \geq 1$

$P(both red) = 5 * P(both blue)$

$\therefore \frac{x}{x+y} * \frac{x-1}{x+y-1} = 5 * \frac{y}{x+y} * \frac{y-1}{x+y-1}$

$x(x-1) = 5y(y-1)$

$P(diff. colours) = 6 * P(both blue)$

$\frac{x}{x+y} * \frac{y}{x+y-1} + \frac{y}{x+y} * \frac{x}{x+y - 1} = 6 * \frac{y}{x+y} * \frac{y-1}{x+y-1}$

$2xy = 6y(y-1)$

$x = 3(y-1) ( y \neq 0 )$

$\therefore x = 3(y-1) , x(x-1) = 5y(y-1)$

$3(y-1)(3(y-1) - 1) = 5y(y-1)$

$(y-1)[3(3(y-1) - 1) - 5y = 0$

$(y-1)(4y - 12) = 0$

$y = 1 , y = 3$
$x = 0 , x = 6$

$\therefore $ number of red balls = x = 6 (since x \neq 0 )$

rand_althor · Oct 7, 2015

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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

sida1049 said:
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.

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

The answers say 1560.

Zlatman · Oct 7, 2015

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

sida1049 · Oct 7, 2015

Re: HSC 2015 3U Marathon

Zlatman said:
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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

Zlatman said:
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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

Here's another probability one:
Two points are picked at random on the unit circle $x^2+y^2=1$ . 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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Here's another probability one:
Two points are picked at random on the unit circle $x^2+y^2=1$ . 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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Here's another probability one:
Two points are picked at random on the unit circle $x^2+y^2=1$ . 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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Here's another probability one:
Two points are picked at random on the unit circle $x^2+y^2=1$ . 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 · Oct 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I got 1/5 ...?

What was your method of selecting the points randomly?

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I got 1/5 ...?

The answer is: 2/3.

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
What was your method of selecting the points randomly?

I just found the segment length with chord 1 and divided that by total length aka circumference.

InteGrand · Oct 7, 2015

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:

InteGrand said:
$Haven't seen the second part you added yet, but for the first one, it's ambiguous because we get different answers depending on our choice of selecting a ``random'' point - it is not specified what method should be used.$

$One method can be we fix one endpoint of the chord, say on the point $(1,0)$ on the unit circle, and then we choose an angle from the uniform distribution on $\left(-180^\circ,180^\circ \right]$, $\mathcal{U}(-180,180)$, and the other endpoint of the chord corresponds to the ray with this angle. It is easy to show that the chord length equals or is less than the radius if and only if the angle picked is between $-60^\circ$ and $60^\circ$. So the chord will be longer than the radius iff we pick the angle in an interval of ``length'' $240^\circ$. Hence the required probability will be $\frac{240}{360}=\frac{2}{3}=0.666...$.$

$Another method would be to say that we will pick a height $h\in [0,1]$ \textit{above} the centre of the circle (from the uniform distribution on $[0,1]$, $\mathcal{U} (0,1)$), and take the chord that corresponds to the horizontal line at this height (i.e. only in the upper semicircle, due to symmetry of the circle). One may show that the chord is longer than the radius iff $0\leq h < \frac{\sqrt{3}}{2}$. Hence the required probability is $\frac{\sqrt{3}}{2}\approx 0.866$.$

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
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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