n(S) = ⁶C₂ = \(\frac{6\times5}{2\times1}\) = 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) = \(\frac{n(A)}{n(S)}\) = \(\frac{14}{15}\)

(b) Let B = sum is at least 8 = {(5, 3) (4, 5)) ⇒ n(B) = 2 

⇒ n(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{2}{15}\)

(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) = \(\frac{n(B)}{n(S)}\) = \(\frac{8}{15}\)