To calculate the probability that the sum of the three numbers on the dice is 11 when three dice are thrown simultaneously, we can break the problem into two parts:

1. Total possible outcomes: Each die has 6 faces, so when three dice are thrown, the total number of possible outcomes is: $6 \times 6 \times 6 = 216$

2. Favorable outcomes: We need to count how many combinations of three numbers from the dice sum up to 11. The three numbers on the dice range from 1 to 6, so we are looking for combinations of three integers $x_1, x_2, x_3$ such that: $1 \leq x_1, x_2, x_3 \leq 6 \quad \text{and} \quad x_1 + x_2 + x_3 = 11$ Let's systematically count the number of favorable outcomes by listing all possible triples:

   • (6, 4, 1)
   • (6, 3, 2)
   • (6, 2, 3)
   • (6, 1, 4)
   • (5, 5, 1)
   • (5, 4, 2)
   • (5, 3, 3)
   • (5, 2, 4)
   • (4, 4, 3)
   • (4, 3, 4)
   • (4, 5, 2)

There are likely several triples that can give