Answer
$P(jury~of~2~students~and~3~faculty)=\frac{3360}{8568}\approx0.3922$
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 2 at a time:
$_{8}C_2=\frac{8!}{2!(8-2)!}=\frac{8!}{2!\times6!}=\frac{8\times7\times6\times5\times4\times3\times2\times1}{2\times6\times5\times4\times3\times2\times1}=\frac{8\times7}{2}=28$
The number of combinations of 10 distinct faculty taken 3 at a time:
$_{10}C_3=\frac{10!}{3!(10-3)!}=\frac{10!}{3!\times7!}=\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times7\times6\times5\times4\times3\times2\times1}=\frac{10\times9\times8}{6}=120$
Using the Multiplication Rule of Counting (page 298):
$N(jury~of~2~students~and~3~faculty)=28\times120=3360$
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):
$P(jury~of~2~students~and~3~faculty)=\frac{N(jury~of~2~students~and~3~faculty)}{N(S)}=\frac{3360}{8568}\approx0.3922$
