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Thermal Statistics - Microstate Probabilities

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Homework Statement



For a system in equilibrium at temperature T, the probability of finding it in a microstate m is:

P(m) = ([itex]1/Z[/itex])exp([itex]-E/kT[/itex])
where Z is the partition function.

There are three accessible microstates, two with energy E[itex]_{a}[/itex] and one with energy E[itex]_{b}[/itex].

Two identical and indistinguishable bosons can occupy these microstates. What's the probability that both these bosons have energy E[itex]_{b}[/itex]?

Homework Equations



I have the values for E[itex]_{a}[/itex] and E[itex]_{b}[/itex].
It asked me to give a formula for Z which I put as Z = [itex]\Sigma[/itex]exp([itex]-E/kT[/itex])
T = 300K
I calculated Z to be 1.08

The Attempt at a Solution



What I've got at the moment is this:
Let the number of particles in states a, a, b be n[itex]_{1}[/itex], n[itex]_{2}[/itex], n[itex]_{3}[/itex]:
n[itex]_{1}[/itex] n[itex]_{2}[/itex] n[itex]_{3}[/itex]
1 1 0
1 0 1
0 1 1
2 0 0
0 2 0
0 0 2

So the probability of both bosons being in state n[itex]_{3}[/itex] aka b would be [itex]1/6[/itex], but this seems way too obvious for 4 marks. I also tried to use the Bose-Einstein distribution but I don't know the chemical potential.

Thanks
 

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