n(S) = 6C2 =
6×5
2×1
= 15
(a) Let A: sum is at most 8 i.e max B
A = {(O, 1) (0, 2) (0,3) (0,4) (0,5) (1,2) (1,3) (1.4) (1,5) (2,3) (2, 4) (2, 5) (3, 4) (3, 5))
n(A)=14
P(A) =
n(A)
n(S)
=
14
15
(b) Let B = sum is at least 8 = {(5, 3) (4, 5)) ⇒ n(B) = 2
⇒ n(B) =
n(B)
n(S)
=
2
15
(c) Let C : Sum is prime number
C = { (0. 1) (0, 2) (0, 3) (0, 5) (1, 2) (1. 4) (2, 3) (3, 4)) ⇒ n(C) = 8
P(C) =
n(B)
n(S)
=
8
15
