Question 22163: 4. When a pair of dice is rolled, the total will range from 2 (1,1) to 12 (6,6). It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over.
Why will some numbers come up more frequently than others?
Each die has six sides numbered from 1 to 6. How many possible ways can a number be rolled? In other words, we can roll (2,3) or (3,2) or (6,1) and so on. What are the total (x,y) outcomes that can occur?
How might you then estimate the percentage of the time a particular number will come up if the dice are rolled over and over?
Once these percentages have been calculated, how might the mean value of the all the numbers thrown be determined?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE MY COMMENTS IN CAPITAL LETTERS
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4. When a pair of dice is rolled, the total will range from 2 (1,1) to 12 (6,6)....OK
It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over. ....YA
Why will some numbers come up more frequently than others? BECAUSE OF 2 DICE AND TOTAL OF TWO NUMBERS SHOWING UP ON 2 DICE BEING TAKEN.SEE THE TABLE BELOW.IT WONT HAPPEN WITH 1 DICE THROW.ALL NUMBERS FROM 1 TO 6 IN THAT CASE ARE EQUALLY LIKELY.
DICE 1 DICE 2 TOTAL DICE 1 DICE 2 TOTAL total FREQ. %
1 1 2 4 1 5 1 0 2.8
1 2 3 4 2 6 2 1 5.6
1 3 4 4 3 7 3 2 8.3
1 4 5 4 4 8 4 3 11.1
1 5 6 4 5 9 5 4 13.9
1 6 7 4 6 10 6 5 16.7
2 1 3 5 1 6 7 6 19.4
2 2 4 5 2 7 8 5 22.2
2 3 5 5 3 8 9 4 25.0
2 4 6 5 4 9 10 3 27.8
2 5 7 5 5 10 11 2 30.6
2 6 8 5 6 11 12 1 33.3
3 1 4 6 1 7
3 2 5 6 2 8
3 3 6 6 3 9
3 4 7 6 4 10
3 5 8 6 5 11
3 6 9 6 6 12
Each die has six sides numbered from 1 to 6. How many possible ways can a number be rolled? In other words, we can roll (2,3) or (3,2) or (6,1) and so on. What are the total (x,y) outcomes that can occur? SEE TABLE ABOVE
How might you then estimate the percentage of the time a particular number will come up if the dice are rolled over and over? SEE TABLE ABOVE FOR PERCENT
Once these percentages have been calculated, how might the mean value of the all the numbers thrown be determined? BY TAKING WEIGHTED AVERAGE OF THE VALUE MULTIPLIED BY FREQUENCY..THAT IS
(1*0+2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)/36=7
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