Question 1196729: Construct a probability distribution for the sum shown on the faces when two dice, each with 7 faces, are rolled.
Found 2 solutions by ikleyn, math_tutor2020: Answer by ikleyn(52803) (Show Source):
You can put this solution on YOUR website! .
This question assumes that you just know on how to do it for the classic 6-faced dice.
Why don't you make it for the 7-faced dice in the same way ?
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Here is what the addition table looks like when we add values from two 7-sided dice
+ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
For example, if we roll a 1 on a blue die and a 7 on a red die, then we get 1+7 = 8 as shown in the upper right corner of that table.
The possible sums are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
X = sum of the two 7-sided dice
P(X) = probability of that sum showing up
The X values range from 2 to 14 inclusive.
The sum "2" shows up exactly once out of 7*7 = 49 possible outcomes. Therefore P(X) = 1/49 when X = 2
X | P(X) | 2 | 1/49 | 3 | | 4 | | 5 | | 6 | | 7 | | 8 | | 9 | | 10 | | 11 | | 12 | | 13 | | 14 | |
Then we have the sum "3" show up twice out of 49 possible outcomes. We write 2/49 next to 3 like so
X | P(X) | 2 | 1/49 | 3 | 2/49 | 4 | | 5 | | 6 | | 7 | | 8 | | 9 | | 10 | | 11 | | 12 | | 13 | | 14 | |
Keep this process going until you have this completed table.
X | P(X) | 2 | 1/49 | 3 | 2/49 | 4 | 3/49 | 5 | 4/49 | 6 | 5/49 | 7 | 6/49 | 8 | 7/49 | 9 | 6/49 | 10 | 5/49 | 11 | 4/49 | 12 | 3/49 | 13 | 2/49 | 14 | 1/49 |
You could write the table out like this if you prefer
X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | P(X) | 1/49 | 2/49 | 3/49 | 4/49 | 5/49 | 6/49 | 7/49 | 6/49 | 5/49 | 4/49 | 3/49 | 2/49 | 1/49 |
A pattern to notice is the numerators start to increase {1,2,3,4,5,6,7} when going from X = 2 to X = 8; afterward we have a decreasing set of numerators {6,5,4,3,2,1}
Optionally you could reduce 7/49 to get 1/7, but I find it's better to have all the denominators be the same.
Side note: all of the P(X) values are between 0 and 1. Also, the P(X) values sum to 1.
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