Answer
$ P(E) = 1.5 \times 10^{-6}$
Work Step by Step
The unoccupied probability state can be found at the top of valence band where E = 0 eV
$ 1 - P(E) = \frac{1}{e^{E_1 - E_F/kT} + 1}$
$ P(E) = 1 - \frac{1}{e^{E_1 - E_F/kT} + 1}$
$ P(E) = \frac{1}{e^{-(E_1 - E_F)/kT} + 1}$
$ P(E) = \frac{1}{e^{-(0 eV - 0.335 eV)/(8.62 \times 10^{-5} eV/K)(290 K)} + 1}$
$ P(E) = \frac{1}{e^{(0.335 eV)/(8.62 \times 10^{-5} eV/K)(290 K)} + 1}$
$ P(E) = 1.5 \times 10^{-6}$
