Answer:
When the two dice are tossed the sample space is:
S: { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2),(3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4),(5,5), (5,6), (6,1), (6,2), (6,3), (6,4) ,(6,5), (6,6) }
Explanation:
a) Let A be an event that the difference between the numbers is 2.
A: { The difference between the numbers is 2 }
A: { (1,3), (2,4), (3,1), (3,5), (4,2), (4,6), (5,3), (6,4)}
P(A) = 8/36 = 4/18 = 2/9
P(A) = 2/9
b)
Let B be the event that a 5 does not appear on either die.
B: { A 5 does not appear on either die }
B: { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4) }
P(B) = 25/36
c)
Let C be the event that the sum of numbers is odd
C: { The sum of the numbers is odd }
C: { (1,2), (1,4), (1,6), (2,1 ), (2,3 ), (2,5 ), (3,2 ), (3,4 ), (3,6 ), (4,1 ), (4,3 ), (4,5 ), (5,2 ), (5,4 ), (5,6 ), (6,1 ), (6,3 ), (6,5 ) }
P(C) = 18/36 = 9/18 = 1/2
P(C) = 1/2