Answer
$P(jury~of~all~students)=\frac{1}{153}\approx0.006536$
Work Step by Step
The order in which the individuals are selected does not matter and no individual can be selected more than once.
The number of combinations of 8 distinct students taken 5 at a time:
$N(jury~of~all~students)=~_8C_5=\frac{8!}{5!(8-5)!}=\frac{8!}{5!\times3!}=\frac{8\times7\times6\times5\times4\times3\times2\times1}{5\times4\times3\times2\times1\times3\times2\times1}=\frac{8\times7\times6}{3\times2\times1}=56$
The number of combinations of 18 distinct individuals (8 students and 10 faculty) taken 5 at a time:
$N(S)=~_{18}C_5=\frac{18!}{5!(18-5)!}=\frac{18!}{5!\times13!}=\frac{18\times17\times16\times15\times14\times13!}{5\times4\times3\times2\times1\times13!}=\frac{1,028,160}{120}=8568$
Using the Classical Method (page 259):
$P(jury~of~all~students)=\frac{N(jury~of~all~students)}{N(S)}=\frac{56}{8568}=\frac{1}{153}\approx0.006536$
