If -3 were one die, then the possible sums of two dies would be -6,-5,-4,-3,-2,-1: avg -3.5
if -2 were one die, then the possible sums of two dies would be -5,-4,-3,-2,-1,0: avg-2.5
the figure all possibilities for the first die, and the avg sum on each. Add all the sums, divide by six. That is the most probable sum.
In my head, I am seeing avg of sum of 1 for all combinations.
This is my second post . I still don't get and it is due tomorrow! !!
Jill made two 6 sided dice and labeled each od them with these numbers
-3, -2, -1, 0, 1, 2
Suppose this pair of dice was tossed 1000 times and the numbers on the two top faces were added. Which sum would probably be the most common? And why??
3 answers
How about this:
create a table with 6 rows and 6 columns
label the rows: -3, -2, -1, 0, 1, 2
and the columns the same way
now fill in the entries by adding the row and column positions.
Look at the sums. Which is the most common and how many times does it happen?
multiply 1000 by that count over 36
create a table with 6 rows and 6 columns
label the rows: -3, -2, -1, 0, 1, 2
and the columns the same way
now fill in the entries by adding the row and column positions.
Look at the sums. Which is the most common and how many times does it happen?
multiply 1000 by that count over 36
In the table I suggested, the most common sum is -1, and it occurs 6 times.
(just like in 2 standard die, the most common sum is 7 and it occurs 6 times)
the prob of that most common sum is 6/36 = 1/6
If we were to toss our dice 1000 times, we could expect a sum of -1 appr
1000(1/6) or 167 times
(just like in 2 standard die, the most common sum is 7 and it occurs 6 times)
the prob of that most common sum is 6/36 = 1/6
If we were to toss our dice 1000 times, we could expect a sum of -1 appr
1000(1/6) or 167 times
1 answer