a) The set of realizations of X can be given by the possible outcomes from the sum of two fair dice: {2,3,4,5,6,7,8,9,10,11,12}
b) If we consider (B,R) a possible outcomes from the blue and red dice, we have the following outcomes for every value for X:
[tex]\begin{gathered} X=2\text{ if}(B,R)\in\mleft\lbrace(1,1)\mright\rbrace \\ X=3\text{ if }(B,R)\in\mleft\lbrace(1,2\mright),(2,1)\} \\ X=4\text{ if }(B,R)\in\mleft\lbrace(1,3\mright),(2,2),(3,1)\} \\ X=5\text{ if }(B,R)\in\mleft\lbrace(1,4),(2,3),(3,2),(4,1)\mright\rbrace \\ X=6\text{ if }(B,R)\in\mleft\lbrace(1,5\mright),(2,4),(3,3),(4,2),(5,1)\}_{} \\ X=7\text{ if }(B,R)\in\mleft\lbrace(1,6\mright),(2,5),(3,4),(4,3),(5,2),(6,1)\} \\ X=8\text{ if }(B,R)\in\mleft\lbrace(2,6),(3,5),(4,4),(5,3),(6,2)\mright\rbrace_{} \\ X=9\text{ if }(B,R)\in\mleft\lbrace(3,6\mright),(4,5),(5,4),(6,3)\} \\ X=10\text{ if }(B,R)\in\mleft\lbrace(4,6\mright),(5,5),(6,4)\} \\ X=11\text{ if }(B,R)\in\mleft\lbrace(5,6\mright),(6,5)\} \\ X=12\text{ if }(B,R)\in\mleft\lbrace(6,6\mright)\} \end{gathered}[/tex]Therefore, the distribution of X, P(X) is given by the ratio between the number of possibilities for a given value of X and the total number of 36 combinations for (B,R)
Then we have:
[tex]\begin{gathered} P(X=2)=\frac{1}{36} \\ P(X=3)=\frac{2}{36}=\frac{1}{18} \\ P(X=4)=\frac{3}{36}=\frac{1}{12} \\ P(X=5)=\frac{4}{36}=\frac{1}{9} \\ P(X=6)=\frac{5}{36} \\ P(X=7)=\frac{6}{36}=\frac{1}{6} \\ P(X=8)=\frac{5}{36} \\ P(X=9)=\frac{4}{36}=\frac{1}{9} \\ P(X=10)=\frac{3}{36}=\frac{1}{12} \\ P(X=11)=\frac{2}{36}=\frac{1}{18} \\ P(X=12)=\frac{1}{36} \end{gathered}[/tex]c)
The expected value for X is given by:
[tex]E\lbrack X\rbrack=\sum ^{12}_{X=2}X\cdot P(X)[/tex]Based on the values obtained in part b, we have:
[tex]E\lbrack X\rbrack\approx6.69[/tex]