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 B has exactly one unit of energy?
Bump
I get 33.333%
The each unit of energy has 3 possible landing. So the total possibilities are 3*3*3 = 27. There are 3 ways in which B has one unit and is the first unit. (BAA, BAC, BCC). So, there must also be 3 ways where B gets the second unit and 3 ways where B gets the third. So 9/27=.33333
To find the probability that molecule B has exactly one unit of energy, we need to consider all the possible arrangements of energy and count the favorable outcomes.
Let's consider the three units of energy as three indistinguishable identical balls and imagine we are distributing them into three labeled boxes representing the molecules A, B, and C.
We can use a counting technique called stars and bars to solve this problem. The three units of energy (balls) can be represented as stars (***), and the two boundaries separating the boxes can be represented as bars (||). For example, the arrangement '*||***' represents the energy distribution where molecule A has no units, molecule B has one unit, and molecule C has two units.
Now let's consider the favorable outcomes where molecule B has exactly one unit of energy. There are three possible ways to arrange one unit of energy between the stars:
1. *|*|**
2. *|**|*
3. **|*|*
For each of these arrangements, we need to count the number of ways to rearrange the remaining two units of energy among the three molecules A and C. This can be done using the concept of permutations.
For the arrangement *|*|**, we have one unit of energy for molecule B and no units for molecules A and C. The remaining two units can be arranged in (2!)/(0!2!) = 1 way.
For the arrangement *|**|*, we have one unit of energy for molecule B, one unit for molecule C, and no units for molecule A. The remaining two units can be arranged in (2!)/(0!2!) = 1 way.
For the arrangement **|*|*, we have one unit of energy for molecule B, one unit for molecule A, and no units for molecule C. The remaining two units can be arranged in (2!)/(0!2!) = 1 way.
Now we sum up the number of favorable outcomes: 1 + 1 + 1 = 3.
Since each of the ten possible arrangements is equally probable, the total number of outcomes is 10.
Therefore, the probability that molecule B has exactly one unit of energy is 3/10 or 0.3 (30%).