Respuesta :
Consider the set of all possible outcomes when rolling a fair die twice,
[tex]S=\lbrace(1,1),(1,2),(1,3),...,(1,6),(2,1),...,(6,5),(6,6)\rbrace\rightarrow\text{ 36 elements}[/tex]There are 36 elements in set S, all equally probable.
A) Notice that
[tex]\begin{gathered} 1=P(X>5)+P(X\leq5) \\ \Rightarrow P(X>5)=1-P(X\leq5) \end{gathered}[/tex]Then, we need to find the elements in S such that their sum is equal to or less than 5,
[tex]\begin{gathered} 1+1\leq5 \\ 1+2\leq5 \\ 2+1\leq5 \\ 3+1\leq5 \\ 1+3\leq5 \\ 3+2\leq5 \\ 2+3\leq5 \\ 4+1\leq5 \\ 1+4\leq5 \\ 2+2\leq5 \end{gathered}[/tex]Those are 10 elements; therefore,
[tex]P(X>5)=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}[/tex]Thus, the probability of event A) is 13/18.
B) Notice that the minimum sum is equal to 2 (1+1) while the maximum sum is 12 (6+6). Then, the possible even numbers one can obtain as the result of a sum are
[tex]2,4,6,8,10,12[/tex]Similarly to part A)
[tex]P(even)=1-P(odd)[/tex]Finding the elements in S whose sum is an odd number,
[tex]\begin{gathered} sum3 \\ 1+2,2+1 \\ sum5 \\ 2+3,3+2,4+1,1+4 \\ sum7 \\ 1+6,6+1,2+5,5+2,3+4,4+3 \\ sum9 \\ 3+6,6+3,4+5,5+4 \\ sum11 \\ 5+6,6+5 \end{gathered}[/tex]Those are 18 elements in total; therefore,
[tex]P(even)=1-\frac{18}{36}=1-\frac{1}{2}=\frac{1}{2}[/tex]The probability of event B is 1/2
