A pair of 8-sides dice have sides numered 1 through 8.Each side has same probability of landing face up.What is the probability that the product of the two numbers on the sides that land face up exceeds36? Why?
I will very appreciate for any suggestion.
Here's a simple way to approach it. Draw up a table. list all the possibilities of the 1st die across the first row, and all the possibilities of the 2nd die down the first column. all the slots in the table represent the combinations of each.
Originally Posted by busyboy
see the table below:
the first row and column are filled in with 1,2,3,4,5,6,7,8 which are the possibilities for each die. the region in the middle is made up of the products of the numbers of the two dice, this table has all possibilies. for example, the case where the first die shows 3 facing up and the second die show 2 facing up is covered by the third row, second column (not counting the headings) the slot here is filled in 6 (which is 3*2).
now we see from the table that there are 8*8 = 64 possible out comes. the ones in bold are the outcomes we desire. these are the ones with products greater than 36, there are 10 such possibilities. so the probability of this occuring is 10/64 = 5/32 = 0.15625
we could also think our way through it. how many combinations are there? well, for each number on die 1, there are 8 possible numbers die 2 could eb, so there are 8*8 = 64 possible outcomes. how many ways can we get a number greater than 36 from the product of two integers between 1 and 8 inclusive.
first way: 6*7
second way: 7*6
third way: 5*8
fourth way: 8*5
note we have to go backwards as well as forwards, that is, if we write down 5*8 we must right down 8*5, since one accounts for if die 1 is 5 and die 2 is 8, while the other accounts for die 1 is 8 and die 2 is 5. doing this you will find there are 10 possibilities and so you put that over the total number of outcomes and you will have your answer