Can someone please help me with this problem ​

There are 36 possibilities in throwing one green and one red die, eg (G1, R1), (G1, R2), etc; (G2, R1), (G2, R2), etc; etc; (G6, R1), (G6, R2), etc.

In the set of possible numbers from throwing a die, there are only 3 possibilities that sum to to 4: (G1, R3), (G2, R2), (G3, R1).

So the answer is 2.3/36 o

2.One-twelfth.

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.