Unlike the Fermi-Dirac distribution function, the Bose-Einstein distribution function $$f(E)=\bar n_r=\frac{1}{e^{\beta(E-\mu)}-1}$$ can be greater than 1, and therefore, doesn't represent a probability. It represents the average number of particles $\bar n_r$ in a single-particle quantum state $r$. What is the expression for a single-particle quantum state $r$ being occupied or unoccupied? Can we related that to $f(E)$ or $\bar n_r$?
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Let $x = e^{-\beta (E - \mu)}$. Essentially by definition, the probability $p_n$ of having occupancy number $n$ $$p_n = \frac{x^n}{Z}$$ where the probability distribution is normalized by the partition function $$Z = 1 + x + x^2 + \ldots = \frac{1}{1-x}.$$ Then the probability that the occupancy number is zero is $$p_0 = \frac{x^0}{Z} = 1 - x.$$ The probability that the occupancy number is nonzero is $x$. The average occupancy is $$\langle n \rangle = \sum_n n p_n = (1-x) \sum_n n x^n = \frac{1}{x^{-1} - 1}$$ which is the result you quoted. In particular, $p_0$ and $\langle n \rangle$ are related by $$p_0 = \frac{1}{1 + \langle n \rangle}.$$
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