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~~~!