Answer:
The probability that the outcome of the roll is an even sum or a sum that is a multiple of 3
P( E∪F) [tex]= \frac{24}{36}[/tex]
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
[tex]P(E) = \frac{n(E)}{n(S)} = \frac{17}{36}[/tex]
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
[tex]P(F) = \frac{n(E)}{n(S)} = \frac{12}{36}[/tex]
n((E∩F) = {(2,4),(3,3),((4,2),(5,1),(6,6)} =5
[tex]P(En F) = \frac{n(En F)}{n(S)} = \frac{5}{36}[/tex]
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)
[tex]\frac{17}{36} +\frac{12}{36} - \frac{5}{36} = \frac{24}{36}[/tex]
