Explanation:

Total possible outcomes when a die is thrown twice = 6×6 = 36 = n( T )
Outcomes for getting a sum 5 from both throws
= { ( 1, 4 ) , ( 2, 3 ) , ( 3, 2 ) , ( 4 1 ) }
n(E) = 4
Thus, P ( E ) = n( E)/n ( T )
= 4/36 = 1/9