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Question 3 c (1 Viewer)

tieki

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For the probability question, yes...the prob of getting at least two sums of five was 1 - P(getting one sum of five) - P(getting no sums of 5)
 

detsiiFLUEY

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q3 (c) (i) Explain why the probability of getting a sum of 5 when one pair of fair dice is tossed is 1/9

(ii) Find the probability of getting a sum of 5 at least twice when a pair of dice is tossed 7 times.

Counting techniques used for the 1st part, Binomial for 2nd part ^_^
 

iambored

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cool! but did you get some weird number, like, for example 12872983/123981029
 
N

ND

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Originally posted by crazylilmonkee
oh shit.. 'at least'
fark i read too fast
Heheh yeh i hate that, i did the same thing in 4u, except i missed the 'at least' bit.
 

failingTheHsc

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for the first part i wrote out all the possible dice rolls and i got probability of getting sum 5/36???

1+4
2+3
3+2
4+1

wat the hell i dunno wat the hell i did but i counted 5
argh
 
N

ND

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Originally posted by iambored
cool! but did you get some weird number, like, for example 12872983/123981029
I dunno, my answer was in decimal (don't remember what it was though).
 

ghoolz

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i used the binomial probability equation that was

P(x)= ncr . p^r . q^(n-r)

then done that twice for the atleast part and subtracted from one but i left it in an unexpanded form. because other wise you get a bad decimal and they dont like that they prefer to leave it unexpanded

i think it ended up being

1 - (8/9)^7 - 7c1 . (1/9) . (8/9)^6

which its decimal equivalent is 0.17788

i think thats it
 

Ragerunner

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Question 3c I wrote a 2-3 page essay. It went something like this but a hell of a lot longer.


"Using an advanced technique, I drew up a digram consisting of rows and column, within those columns and rows lie numbers. The numbers on the 1st row consist of the number 1,2,3,4,5,6. The numbers on the columns row on the left consist of 1,2,3,4,5,6. These numbers represent the dice outcomes. The numbers inside the diagram show the sum of these numbers. As outputting I had the results of 1+1, 2+1, 3+1, 4+1, 5+1 etc.... all the way to the end. Using these results the number of possible overall outcomes were counted and resulted in 36. Looking at the diagram, the sum of 2 of the numbers that equalled 5 were observed and resulting in 4 being found. As a probability blah blah blah it goes on."

I didn't care about 3U, might as well guarantee I score 1 mark :D
 
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ND

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Originally posted by Ragerunner

"Using an advanced technique, I drew up a digram consisting of rows and column, within those columns and rows lie numbers. The numbers on the 1st row consist of the number 1,2,3,4,5,6. The numbers on the columns row on the left consist of 1,2,3,4,5,6. These numbers represent the dice outcomes. The numbers inside the diagram show the sum of these numbers. As outputting I had the results of 1+1, 2+1, 3+1, 4+1, 5+1 etc.... all the way to the end. Using these results the number of possible overall outcomes were counted and resulted in 36. Looking at the diagram, the sum of 2 of the numbers that equalled 5 were observed and resulting in 4 being found. As a probability blah blah blah it goes on."
Hahah, that kinda answer is not required in teh HSC maths course yet.
 

iambored

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Originally posted by Ragerunner
Question 3c I wrote a 2-3 page essay. It went something like this but a hell of a lot longer.


"Using an advanced technique, I drew up a digram consisting of rows and column, within those columns and rows lie numbers. The numbers on the 1st row consist of the number 1,2,3,4,5,6. The numbers on the columns row on the left consist of 1,2,3,4,5,6. These numbers represent the dice outcomes. The numbers inside the diagram show the sum of these numbers. As outputting I had the results of 1+1, 2+1, 3+1, 4+1, 5+1 etc.... all the way to the end. Using these results the number of possible overall outcomes were counted and resulted in 36. Looking at the diagram, the sum of 2 of the numbers that equalled 5 were observed and resulting in 4 being found. As a probability blah blah blah it goes on."

I didn't care about 3U, might as well guarantee I score 1 mark :D
HAHAHA!!! didn't u know english was over?? nah, that answer's gold, i do that when i don't know what to do, you never know how they mark
 

iambored

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Originally posted by Ragerunner
Question 3c I wrote a 2-3 page essay. It went something like this but a hell of a lot longer.


"Using an advanced technique, I drew up a digram consisting of rows and column, within those columns and rows lie numbers. The numbers on the 1st row consist of the number 1,2,3,4,5,6. The numbers on the columns row on the left consist of 1,2,3,4,5,6. These numbers represent the dice outcomes. The numbers inside the diagram show the sum of these numbers. As outputting I had the results of 1+1, 2+1, 3+1, 4+1, 5+1 etc.... all the way to the end. Using these results the number of possible overall outcomes were counted and resulted in 36. Looking at the diagram, the sum of 2 of the numbers that equalled 5 were observed and resulting in 4 being found. As a probability blah blah blah it goes on."

I didn't care about 3U, might as well guarantee I score 1 mark :D
HAHAHA!!! didn't u know english was over?? nah, that answer's gold, i do that when i don't know what to do, you never know how they mark
 

Ragerunner

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I wrote this on my site.

Check it out for my exam thoughts. http://unix.ozclans.com

ALL THIS FOR 1 MARK! YAY

This is the closest to what I wrote for that dice question.
I wrote the entire exam paper into my booklet.

Question was Explain why the probability of getting a sum of 5 when one pair of dice is tossed is 1/9.

My answer:

"Using an advanced technique, I drew up a lattice diagram consisting of rows and column, within those columns and rows lie numbers. The numbers on the 1st row consist of the numbers 1,2,3,4,5,6. The numbers on the columns row on the left consist of 1,2,3,4,5,6. These numbers represent all the possible dice outcomes of a fair dice with 6 sides labelled 1 to 6 with all equal side lengths and equally balanced dots on each face so as no bias can occur. The numbers inside the lattice diagram show the sum of these numbers. As outputting I had the results of 1+1, 2+1, 3+1, 4+1, 5+1 6+1, 1+2, 2+2, 3+2, 4+2, 5+2, 6+2, 1+3, 2+3, 3+3, 4+3,5+3, 6+3, 1+4, 2+4, 3+4, 5+4, 6+4, 1+5, 2+5, 3+5, 4+5, 5+5, 6+5, 1+6, 2+6, 3+6, 4+6, 5+6, and 6+6.

Using these results I added them together and it showed the sum as 2,3,4,5,6,3,4,5,6,7,8,4,5,6,7,8,9,10,5,6,7,8,9,10,11,6,7,8,9,10,11 Upon observing these sums that have just been added using a linear search I went through each number in the lattice and identified the one named 5 and kept count on how many 5s I encounted. Using this lattice diagram there were 4 numbers that had the sum of 5, and thus by using the probability rule of number of outcomes divided by total outcomes I came up with a number of 4/36. When using an advanced technique of simplification, I cancelled out the 4/36 to 1/9, but divided 4 by 4 =1 and 36 divided by 4 = 9.

And thus the magical number appeared as 1/9 which is the answered required and the explanation has just been given."

This better be on the 2004 STANDARDS PACKAGE. SIFN'T EXEMPLAR~~~!
 

Constip8edSkunk

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Originally posted by Ragerunner
...
When using an advanced technique of simplification,
...
LMAO!!! hahahahaha :rofl::rofl::rofl:

pure gold!



<10 years from now, the HSC students will look back and wonder what the fuss was about, as they routinely memorise prepared essays for the 40min piece in q7...and sif only 250 words will get exemplar>
:D
 

Ragerunner

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i did a bit more than that i think.

i also drew diagrams to explain it to take up space

i.e.

 

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