Consider a system of three molecules A, B, and C. Suppose that three units of energy can be distributed over the three atoms. Each atom can have no energy, one unit of energy, two units of energy, or all three units of energy. Assume that each of the ten possible arrangements of the three units of energy is equally probable. What is the probability that molecule C has some energy?

To determine the probability that molecule C has some energy, we first need to find the total number of equally probable arrangements and then count how many of those arrangements have molecule C with some energy.

Let's consider the possible energy configurations for the three molecules:

- No energy for any of the molecules (0-0-0)
- 3 units of energy for molecule A, and none for the others (3-0-0)
- 2 units for A and 1 unit for B, no energy for C (2-1-0)
- 2 units for A and 1 unit for C, no energy for B (2-0-1)
- 1 unit for A and 2 units for B, no energy for C (1-2-0)
- 1 unit for A and 2 units for C, no energy for B (1-0-2)
- 1 unit for B and 2 units for A, no energy for C (2-1-0)
- 1 unit for B and 2 units for C, no energy for A (0-2-1)
- 1 unit for C and 2 units for A, no energy for B (2-0-1)
- 1 unit for C and 2 units for B, no energy for A (0-1-2)
- All 3 units for molecule C (0-0-3)

There are 11 different configurations in total.

Out of these, we can see that there are 6 configurations where molecule C has some energy: 2-0-1, 2-1-0, 0-2-1, 0-1-2, 1-0-2, and 1-2-0.

Therefore, the probability that molecule C has some energy is given by the number of favorable outcomes (6) divided by the total number of equally probable outcomes (11):

Probability = 6/11 ≈ 0.545 or 54.5%