Help PLEASE! Two distinct number cubes are rolled together. Each number cube has sides numbered 1 through 6.

What is the probability that the outcome of the roll is an even sum or a sum that is a multiple of 3?

Enter your answer, in simplest fraction form, in the box.

The probability that the outcome of the roll is an even sum or a sum that is a multiple of 3

P( E∪F)  

Step-by-step explanation:

Given Each number cube has sides numbered 1 through 6

Two distinct number cubes are rolled together

Total number of exhaustive cases n(S) = 36

Let 'E' be the event of getting sum is even

n(E) = {(1,3),(1,5), (2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4), (4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)} = 17

Let 'F' be the event of getting sum that is a multiple of '3'

n(F) = {(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6) = 12

n((E∩F) = {(2,4),(3,3),((4,2),(5,1),(6,6)} =5

The probability that the outcome of the roll is an even sum or a sum that is a multiple of 3

P( E∪F) = P(E) + P(F) - P( E∩F)

should have been evaluated with one odd value and one even value. the expressions have equivalent values when . when , the first expression has a value of 3 and the second

step-by-step explanation: