I really need help with this Algebra 2 problem?

Question

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events: Event A : The sum is greater than 6 . Event B : The sum is an even number. Write your answers as exact fractions

Answer 1

for the first roll of the die you can get 1, 2, 3, 4, 5, or 6

for the second roll of the die you can get 1, 2, 3, 4, 5, or 6

the sums for the first roll plus the second roll are:

1+1=2, 1+2=3, 1+3=4, 1+4=5, 1+5=6, 1+6=7                    

2+1=3, 2+2=4, 2+3=5, 2+4=6, 2+5=7, 2+6=8

3+1=4, 3+2=5, 3+3=6, 3+4=7, 3+5=8, 3+6=9

4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10

5+1=6, 5+2=7, 5+3=8, 5+4=9, 5+5=10, 5+6=11

6+1=7, 6+2=8, 6+3=9, 6+4=10, 6+5=11, 6+6=12

there are 36 possible outcomes

Event A: 21/36=7/12

Event B: 18/36=1/2

 

Answer 2

Event A: The sum is greater than 6.

Possible outcomes: 7,8,9,10,11,12

 

Probability of 12 {6,6}= (1/6)*(1/6)=1/36     

Probability of 11 {5,6}{6,5}= (1/6)*(1/6) +(1/6)*(1/6)= 2/36

Probability of 10 {5,5}{4,6}{6,4}= (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6) = 3/36

Probability of 9 {3,6}{6,3}{4,5}{5,4}= 4/36

Probability of 8 {2,6}{6,2}{3,5}{5,3}{4,4}= 5/36

Probability of 7 {1,6}{6,1}{2,5}{5,2}{3,4}{4,3}= 6/36

 

Probability of event A = (1/36)+(2/36)+(3/36)+(4/36)+(5/36)+(6/36) = 21/36 =7/12

 

 

 

Event B: The sum is an even number

Possible outcomes: 2,4,6,8,10,12

 

Probability of 2 {1,1} = 1/36

Probability of 4 {2,2}{1,3}{3,1}=3/36

Probability of 6 {1,5}{5,1}{4,2}{2,4}{3,3}=5/36

Probability of 8 {2,6}{6,2}{5,3}{3,5}{4,4}=5/36

Probability of 10 {5,5}{4,6}{6,4}=3/36

Probability of 12 {6,6}= 1/36

 

Probability of Event B: (1/36)+(3/36)+(5/36)+(5/36)+(3/36)+(1/36)= 18/36 = 1/2