HSC 2015 MX1 Marathon (archive) (2 Viewers)

Fizzy_Cyst · Mar 19, 2015

Re: HSC 2015 3U Marathon

I get an answer of 0.231..

Don't know if my thought processes are intact though

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

Fizzy_Cyst said:
I get an answer of 0.231..

Don't know if my thought processes are intact though

I believe this is the correct answer.

My calculation:

6P2 × 6C2 × 4! / (6⁶)

Drsoccerball · Mar 19, 2015

Re: HSC 2015 3U Marathon

braintic said:
I believe this is the correct answer.

My calculation:

6P2 × 6C2 × 4! / (6⁶)

I got what Ekman got notsure it sounds right.
So theres 6 courses A,B,C,D,E,F let A be the excluded one therefore to pick any of the dishes 5/6 you multiply by 4/6 for the remaining amount of dishes chooseable, multiply by 3/6 2/6 1/6 and then finally 5/6 since they can pick any of the 5 dishes how is this wrong

Ekman · Mar 19, 2015

Re: HSC 2015 3U Marathon

braintic said:
I believe this is the correct answer.

My calculation:

6P2 × 6C2 × 4! / (6⁶)

Can you explain your thought process pls?

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

Ekman said:
Can you explain your thought process pls?

Assign diners to meals (not meals to diners).

The only way this scenario can occur is if 4 meals are chosen by exactly 1 person, 1 meal is chosen by 2 people, 1 meal is chosen by no-one.

Select the meals which are to be chosen by 0 or 2 people ... 6P2 ... order matters.

Select the two diners who are to assigned to the meal which you have picked for two diners ... 6C2

Arrange the other 4 diners amongst the other 4 meals ... 4!

Total number of ways the meals can be chosen ... 6^6

So 6P2 times 6C2 times 4! / (6^6)

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I got what Ekman got notsure it sounds right.
So theres 6 courses A,B,C,D,E,F let A be the excluded one therefore to pick any of the dishes 5/6 you multiply by 4/6 for the remaining amount of dishes chooseable, multiply by 3/6 2/6 1/6 and then finally 5/6 since they can pick any of the 5 dishes how is this wrong

I'm having difficulty understanding your explanation. The first 5/6 seems to be randomly placed in the middle of a sentence ... I can't work out what you are doing with it. And you will have to explain who 'they' is ... do you mean the last person making a choice ... do you mean 'anyone'?

Drsoccerball · Mar 19, 2015

Re: HSC 2015 3U Marathon

braintic said:
I'm having difficulty understanding your explanation. The first 5/6 seems to be randomly placed in the middle of a sentence ... I can't work out what you are doing with it. And you will have to explain who 'they' is ... do you mean the last person making a choice ... do you mean 'anyone'?

The probability of not picking A is 5/6 then the next probability is 4/6 (The probability of not picking A or B) ect... but since there's one remaining and it can be any of B,C,D,E,F you multiply again by 5/6
EDIT: anyone

VBN2470 · Mar 19, 2015

Re: HSC 2015 3U Marathon

braintic said:
Assign diners to meals (not meals to diners).

The only way this scenario can occur is if 4 meals are chosen by exactly 1 person, 1 meal is chosen by 2 people, 1 meal is chosen by no-one.

Select the meals which are to be chosen by 0 or 2 people ... 6P2 ... order matters.

Select the two diners who are to assigned to the meal which you have picked for two diners ... 6C2

Arrange the other 4 diners amongst the other 4 meals ... 4!

Total number of ways the meals can be chosen ... 6^6

So 6P2 times 6C2 times 4! / (6^6)

You seem to have the right idea, but note that each diner chooses only one meal each, you can't have 4 meals for one diner. I'll post up my solution soon.

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
You seem to have the right idea, but note that each diner chooses only one meal each, you can't have 4 meals for one diner. I'll post up my solution soon.

My solution doesn't allow for 4 meals per diner. The fact that I am assigning diners to meals precludes that possibility. Each diner gets the one meal they are assigned to.

VBN2470 · Mar 19, 2015

Re: HSC 2015 3U Marathon

Ah, I see what you have done, your explanation seemed a bit dodge so I misinterpreted what you said, your answer is correct

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

Ekman said:
$\frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \times \frac{5}{6} = 0.0129$

This is the probability that one SPECIFIC meal is not chosen, and one SPECIFIC person has the same meal as someone else.

braintic · Mar 19, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
Ah, I see what you have done, your explanation seemed a bit dodge so I misinterpreted what you said, your answer is correct

"Dodge" in what way?

VBN2470 · Mar 19, 2015

Re: HSC 2015 3U Marathon

ALTERNATE WAY:

$1. The number of ways in which ONE meal is NOT chosen is ${6 \choose 1}$ \\ 2. The number of ways in which one meal is selected exactly twice is ${5 \choose 1}$. \\ 3. The total number of ways in which the diners can be assigned to the meals is $\frac{6!}{2!}$. \\ 4. Total number of possibilities is $6^6$. \\ Therefore, the probability is $\frac{{6 \choose 1}\times{5 \choose 1}\times\frac{6!}{2!}}{6^6}$ = $\frac{25}{108}$.$

ac_student · Mar 22, 2015

Re: HSC 2015 3U Marathon

New Question:
To estimate the quality of mass-produced items, a customer uses the following double-sampling scheme.
A random sample of ten is inspected. If no defectives are found, the customer accepts the lot. If four or
more defectives are found, the customer rejects the lot. But if one, two or three defectives are found, the
customer then takes a second sample of ten, and if the total of defectives in the two samples together is four
or more, the customer rejects the lot, but otherwise the customer accepts the lot.
Calculate the probability that a lot which is known to be 10% defective is accepted.

braintic · Mar 22, 2015

Re: HSC 2015 3U Marathon

ac_student said:
New Question:
To estimate the quality of mass-produced items, a customer uses the following double-sampling scheme.
A random sample of ten is inspected. If no defectives are found, the customer accepts the lot. If four or
more defectives are found, the customer rejects the lot. But if one, two or three defectives are found, the
customer then takes a second sample of ten, and if the total of defectives in the two samples together is four
or more, the customer rejects the lot, but otherwise the customer accepts the lot.
Calculate the probability that a lot which is known to be 10% defective is accepted.

[part b] Given that the batch is accepted, find the probability that more than 10% of the batch is defective

leehuan · Apr 14, 2015

Re: HSC 2015 3U Marathon

Is this binomial probability?

$For\quad a\quad random\quad pick,\quad chances \quad are \quad 0.1\quad being\quad defective\quad and\quad 0.9\quad being\quad good.\\ Therefore,\quad P(k\quad defective\quad out\quad of\quad 10)\\ =\left( \begin{matrix} 10 \\ k \end{matrix} \right) { \left( 0.1 \right) }^{ k }{ \left( 0.9 \right) }^{ 10-k }$

http://i1164.photobucket.com/albums/q566/Rui_Tong/IMG_1906_zpsalw1ftcr.jpg
Tree diagram.
For part (a), the required pathways to be added are P(0) + P(1,< 3) + P(2, < 2) + P(3, 0)

That's what I got. Shortcut method...?

photastic · Apr 14, 2015

Re: HSC 2015 3U Marathon

Can this question be done without integration?

A Fever is responsible for a rise in temperature of a body of 1C per hour, if no cooling took place. However an adult currently measuring 40C is placed in a 12C cool room which cools the temperature of the body according to Newton’s rate of cooling.
If cooling stops when the body reaches 37C, Calculate when the body will reach 38C.

Carrotsticks · Apr 14, 2015

Re: HSC 2015 3U Marathon

Yes, if you state the exponential model straight off versus deriving it from the first order ode

SpiralFlex · Apr 14, 2015

Re: HSC 2015 3U Marathon

$\textup{Find the}\; n\textup{-th term of the sequence}$

$\frac{1}{10}, \frac{9}{30}, \frac{24}{70}, \frac{46}{150}, \frac{75}{310}, \frac{111}{630} ...$

$\textup{Please only answer if you're doing 3U or if you know how to do it give a slight hint to those}$

InteGrand · Apr 14, 2015

Re: HSC 2015 3U Marathon

SpiralFlex said:
$\textup{Find the}\; n\textup{-th term of the sequence}$

$\frac{1}{10}, \frac{9}{30}, \frac{24}{70}, ...$

$Can be many things probably.$

$One way is to make the denominator $D_n=10+20(n-1)=20n-10$ for $n=1,2,3,\cdots$, and to fit the numerators to a quadratic $N_n=an^2 + bn +c$.$

Edit: not sure if spoilers work properly with LaTeX.

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