three dice are numbered 1 to 6 two of them are red and one is blue. all three dice are rolled.?

Total number of permutations of throwing two red dice = 6 × 6 × 6 = 216

When score of blue dice = 1 :
0 suitable permutation for red dice.

When score of blue dice = 2 :
1 suitable permutation for red dice : (1, 1)

When score of blue dice = 3 :
2 suitable permutations for red dice : (1, 2), (2.1)

When score of blue dice = 4 :
3 suitable permutations for red dice : (1, 3), (2, 2), (3, 1)

When score of blue dice = 5 :
4 suitable permutations for red dice : (1, 4), (2, 3), (3, 2), (4, 1)

When score of blue dice = 6 :
5 suitable permutations for red dice : (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

The required probability
= (0 + 1 + 2 + 3 + 4 + 5)/216
= 15/216

Find the probability of each possible arrangement that does that

I will show them as:
Blue = red + red
1 = no combo P = 1/6 = 36/216
2 = 1 + 1. P = 1/6 x 1/36 = 1/216
3 = 1 + 2. P = 1/6 x 2/36 = 2/216
4 = 1 + 3 or 2 + 2. P = 1/6 x 3/36 = 3/216
5 = 1 + 4 or 2 + 3. P = 1/6 x 4/36 = 4/216
6 = 1 + 5 or 2 + 4 or 3 + 3. P = 1/6 x 5/36 = 5/216
Total is (1+2+3+4+5)/216 = 13/216 = 0.060185185...

Sample space of blue die = {1, 2, 3, 4, 5, 6} (6 outcomes)
Sample space of two red dice = {(1,1), (1,2), ......, (6, 6)} (36 outcomes)
Total outcomes of red and blue dice are 42
The favorable outcomes are
1+1
1+2, 2+1
1+3, 3+1
1+4, 4+1
1+5, 5+1
9 outcomes are favorable

P(sum of no. of two red dice = no. on blue die) = 9/42 = 3/14

PS: The book may also consider 36 as total no. of outcomes, then reqd probability will be 9/36 = 1/4