1)
Two dices can be thrown in 6x6=36 ways
n(S)= 36
Let A denotes the event of getting even number as sum
means sum as , 2,4,6,8,10 12
Hence A={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) , (2,6 ), ( 4,4 ), (4,6 ), ( 4,2 ),
(5,3 ),( 3,5 ),( 6,6 ), (6,4 ),.( 5,1 ), (3,1 ) ,( 6,2 ),( 3,3 ), (5,5 ) },
n(A)= 18
thus,required probability= P(A)= n(A)/n(S)= 18/36= 1/2
2)
Two dices can be thrown in 6x6=36 ways
n(S)= 36
Let A denotes the event of getting odd number as sum
means sum as , 3,5,7,9,11
Hence A={( 1,2 ),(2,1 ) (1,4 ) (4,1 ) , (2,3 ) , (3,2 ), ( 1,6 ), (6,1 ), ( 5,2 ),
(2,5 ),( 3,4 ),( 4,3 ), (6,3 ),( 3,6 ), (4,5 ) ,( 5,4 ),( 6,5 ), (5,6 ) },
n(A)= 18
thus,required probability= P(A)= n(A)/n(S)= 18/36= 1/2