Hi can anybody help me understand how to solve Event B. Can you be very detailed please!!

Question

Can anybody help me with event B because im so confused on how do you get the sum of the divisible number. 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 9. Event B: The sum is divisible by 6. Write your answers as exact fractions.

Answer 1

Put all of the possible outcomes in a table [note: WyzAnt doesn't like multiple spaces).

.....Roll.1.......1......2.....3.....4.....5.....6

Roll 2 ------------------------------------------------

....1................2......3.....4.....5.....6.....7

....2................3......4.....5.....6.....7.....8

....3................4.....5......6.....7.....8.....9

....4................5.....6......7.....8.....9.....10

....5................6.....7......8.....9....10.....11

....6................7.....8......9.....10....11....12

Event A: There are 6 out of 36 cases in which the sum is greater than 9. P(A) = 6/36 = 1/6

Event B: There are 6 out of 36 times that the sum is evenly divisible by 6 (that is, no remainder).

P(B) = 6/36 = 1/6

Answer 2

The possible sums are 2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12.

What is the probability of rolling a sum greater than 6?