Consider a system of 3 molecules, A, B, and C. Suppose the 3 units of energy can be distributed over the three atoms where each atom can have no energy, one unit, two units or 3 units of energy. Assume each of the 10 possible arrangements is equally probable. What's probability that molecule C has some energy?

Out of 10 combinations, three will have C with some energy.

Pr(3/10)

4/10!!!

To find the probability that molecule C has some energy, we need to determine the number of favorable outcomes (where molecule C has some energy) and the total number of possible outcomes.

Let's first list all the possible energy distributions for the three molecules A, B, and C:

1. (0, 0, 3)
2. (0, 1, 2)
3. (0, 2, 1)
4. (0, 3, 0)
5. (1, 0, 2)
6. (1, 1, 1)
7. (1, 2, 0)
8. (2, 0, 1)
9. (2, 1, 0)
10. (3, 0, 0)

Out of these 10 possible arrangements, we can see that in 6 of them, molecule C has some energy (arrangements 1-4 and 6-7).

Therefore, the number of favorable outcomes is 6.

Since each of the 10 possible arrangements is equally probable, the total number of possible outcomes is 10.

To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes
= 6 / 10
= 0.6

Therefore, the probability that molecule C has some energy is 0.6 or 60%.