Answer
$P(1~contains~diet~soda)=\frac{27}{55}\approx0.4909$
Work Step by Step
Combinations of 12 distinct cans (regular or diet soda) taken 3 at a time (the order in which the cans are selected does not matter):
$N(S)=~_{12}C_3=\frac{12!}{3!(12-3)!}=\frac{12!}{3!\times9!}=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times9\times8\times7\times6\times5\times4\times3\times2\times1}=220$
Combinations of 3 distinct diet soda taken 1 at a time (the order in which the cans are selected does not matter):
$N(1~diet~soda~among~the~3~diet~soda)=~_{3}C_1=\frac{3!}{1!(3-1)!}=\frac{3!}{2!\times1!}=\frac{3\times2\times1}{2\times1\times1}=3$
Combinations of 9 distinct regular soda taken 2 at a time (the order in which the cans are selected does not matter):
$N(2~regular~soda~among~the~9~regular~soda)=~_{9}C_2=\frac{9!}{2!(9-2)!}=\frac{9!}{2!\times7!}=\frac{9\times8\times7\times6\times5\times4\times3\times2\times1}{2\times1\times7\times6\times5\times4\times3\times2\times1}=36$
Using the Multiplication Rule of Counting (page 298):
$N(1~diet~soda~and~2~regular~soda)=3\times36=108$
Using the Classical Method (page 259):
$P(1~contains~diet~soda)=\frac{N(1~diet~soda~and~2~regular~soda)}{N(S)}=\frac{108}{220}=\frac{27}{55}\approx0.4909$
