 Solution

Choose the ground-state energy as zero and write $x=\epsilon /(k_{B}T)$. The molecular partition function must sum individual states, so the triply degenerate excited level contributes three Boltzmann factors:The probabilities of occupying the two levels areEach of the three individual excited states has probability $e^{-x}/q$. At high temperature the four states become equiprobable, so the two levels have probabilities $1/4$ and $3/4$, not $1/2$ each. This is the distinction between a level and its state degeneracy.