There are many possible combinations to consider.
Draw a possibility space to find all the outcomes then we decide how many we want
Dice B:
6 sum is:#color(white)(xx)7color(white)(xxx)8color(white)(xxx)9color(white)(xxx)10color(white)(xxx)11color(white)(xxx)12#
5 sum is#color(white)(x.x)6color(white)(xxx)7color(white)(xxx)8color(white)(x.xx)9color(white)(xxx)10color(white)(xxx)11#
4 sum is:#color(white)(xm)5color(white)(x.x)6color(white)(xxx)7color(white)(xx.x)8color(white)(x.xx)9color(white)(xx.x)10#
3 sum is:#color(white)(xx)4color(white)(xxx)5color(white)(xxx)6color(white)(xx.x)7color(white)(xx.x)8color(white)(xxx.)9#
2 sum is:#color(white)(xx)3color(white)(xxx)4color(white)(xxx)5color(white)(xx.x)6color(white)(xx.x)7color(white)(xx.x)8#
1 sum is:#color(white)(..x)2color(white)(x.x)3color(white)(x..x)4color(white)(x.xx)5color(white)(xx.x)6color(white)(xx.x)7#
Dice A:#color(white)(xxx)1color(white)(....x)2color(white)(x...x)3color(white)(x...x)4color(white)(x.x.x)5color(white)(xx.x)6#
There are 36 outcomes from 2 dice.
18 are odd, 18 are even. This can be confirmed by counting the even outcomes in the array above.
In addition to the 18 even numbers there are 2 odd numbers less than 5: 3 and 3.
Therefore of the 36 outcomes there are 20 that are favourable:
#" Probability " = 20/36 = 5/9#
