Question

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?



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Answer 1

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

P( E∪F)
`= (24)/(36)`

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

`P(E) = (n(E))/(n(S)) = (17)/(36)`

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

`P(F) = (n(E))/(n(S)) = (12)/(36)`

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

`P(En F) = (n(En F))/(n(S)) = (5)/(36)`

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)

`(17)/(36) +(12)/(36) - (5)/(36) = (24)/(36)`