If two dice are rolled total sample space
=(6)2=36 Sum of face values can be a composite number.
∴ Sum
=4,6,8,9,10,12 Number of favourable cases
⇒(1,3),(3,1),(2,2),(3,3),(2,4),(4,2),(5,1),(1,5), (2,6),(6,2),(5,3),(3,5),(4,4),(3,6),(6,3),(4,5), (5,4),(4,6),(6,4),(5,5)(6,6)=21
Hence, required probability