In total, there are (\(6^3 = 216\)) combinations.

The minimum sum on the 2 blue die is 2(1+1).

Since we have been asked the probability of red die's sum

to be greater than the blue die's, the minimum number we can have on red die is 3. (1 combination)

When we get a 4 on the red die, there are 3 combinations for the blue die{(1,2),(2,1),(1,1)}

When we get a 5 on the red die, there are 6 combinations for the blue die{(1,2),(2,1),(1,1),(2,2),(1,3),(3,1)}

When we get a 6 on the red die, there are 10 combinations for the blue die{(1,2),(2,1),(1,1),(2,2),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1)}

The probability that the number on the red die will be greater than the sum of the two blue dice = \(\frac{1+3+6+10}{216} = \frac{20}{216} = \frac{5}{54}\)

(Option A)
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