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?
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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
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
1 answer