Question

Two fair dice, one blue and one red, are thrown. Let X(w) bethe random variable that assigns each possible outcome the the sum ofthe two sides facing up. E.g. if the blue die shows 1 and the red oneshows 3, then X assigns this outcome the number 1 + 3 = 4.

Answer 1

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:

`\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}`

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:

`\begin{gathered} P(X=2)=(1)/(36) \\ P(X=3)=(2)/(36)=(1)/(18) \\ P(X=4)=(3)/(36)=(1)/(12) \\ P(X=5)=(4)/(36)=(1)/(9) \\ P(X=6)=(5)/(36) \\ P(X=7)=(6)/(36)=(1)/(6) \\ P(X=8)=(5)/(36) \\ P(X=9)=(4)/(36)=(1)/(9) \\ P(X=10)=(3)/(36)=(1)/(12) \\ P(X=11)=(2)/(36)=(1)/(18) \\ P(X=12)=(1)/(36) \end{gathered}`

c)

The expected value for X is given by:

`E\lbrack X\rbrack=\sum ^(12)_(X=2)X\cdot P(X)`

Based on the values obtained in part b, we have:

`E\lbrack X\rbrack\approx6.69`