Answer
$ P(E) = 4.78 \times 10^{-10} $
Work Step by Step
The probability of occupied state with energy E is
$ P(E) = \frac{1}{e^{E_1 - E_F/kT} + 1}$
For pure Silicon, the energy gap is 1.11 eV, hence pure silicon has half of the energy gap value. The equation becomes
$ P(E) = \frac{1}{e^{E_{Gap}/2kT} + 1}$
At
$T= 300 K$
$KT = (8.62 \times 10^{-5} eV/k) (300K)$
$KT = 0.02586eV$
Solve for P(E)
$ P(E) = \frac{1}{e^{1.11 eV/2 (0.02586eV)} + 1}$
$ P(E) = 4.78 \times 10^{-10} $
