When a fair die is rolled, it's probability of landing on any side is the same.
To calculated the probability P(A) and P(B), Let us write out the possible outcomes of the event that the die was rolled twice. In each roll we can have the possible outcome of 1 to 6. The set of possible outcomes for the two rolls are shown below.
[tex]\lbrace(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,5),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}[/tex]The above are the possible outcomes of the two rolls of die.
The sum of each outcome are;
[tex]\left\lbrace 2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\right\rbrace [/tex]To get the value of P(A), Let's count the outcomes that have their sum greater than 6 and divide by the total;
[tex]\begin{gathered} P(A)=\frac{n(A)}{n(T)}=\frac{21}{36} \\ P(A)=\frac{21}{36} \end{gathered}[/tex]Then for P(B), Let's count the outcomes that have their sum even and divide by the total;
[tex]\begin{gathered} P(B)=\frac{n(B)}{n(T)}=\frac{18}{36} \\ P(B)=\frac{18}{36} \end{gathered}[/tex]Therefore;
[tex]\begin{gathered} P(A)=\frac{21}{36} \\ P(B)=\frac{18}{36} \end{gathered}[/tex]