Consider the table
[tex]\begin{array}{ccccccc}\text{Numbers on cubes} & 1 & 2 & 3 & 4 & 5 & 6\\1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \\ \end{array}[/tex]
where pair (i,j) means that on first cube is number i and on second cube is number j.
Now you can find the number of odd sums from all 36 sums:
- 1 sum 2 (for pair (1,1) the sum is 1+1=2);
- 2 sums 3 (for pairs (1,2) and (2,1) the sum is 3);
- 4 sums 5 (for pairs (1,4), (2,3), (3,2), (4,1));
- 6 sums 7 (for pairs (1,6), (2,5), (3,4), (4,3),(5,2),(6,1));
- 2 sums 11 (for pairs (5,6), (6,5)).
In total there are 1+2+4+6+2=15 prime sums.
Then the probability of the sum of the cubes being prime is
[tex]Pr=\dfrac{15}{36}=\dfrac{5}{12}\approx 0.417.[/tex]
