The first step is calculating the sample space S= {2, 3, ,3 ,4,4,4 ,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12}
count the frequency of each possible outcome and divide with total count n =36 to find PDF then calculate CDF for each X by adding the current and all preceding PDFs.
X PDF CDF
2 "\\frac 1 {36}" "\\frac 1 {36}"
3 "\\frac 1 {18}" "\\frac {1} {12}"
4 "\\frac 1 {12}" "\\frac 16"
5 "\\frac 19" "\\frac 5 {18}"
6 "\\frac 5 {36}" "\\frac 5 {12}"
7 "\\frac 1 6" "\\frac 7 {12}"
8 "\\frac 5 {36}" "\\frac {13} {18}"
9 "\\frac 1 9" "\\frac 5 6"
10 "\\frac 1 {12}" "\\frac {11} {12}"
11 "\\frac 1 {18}" "\\frac {35} {36}"
12 "\\frac 1 {36}" 1
The most likely sum is 7 it has the highest PDF .
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