Statistics: Informed Decisions Using Data (4th Edition)

$P(both~cans~contain~diet~soda)=\frac{1}{22}\approx0.04545$
- First can: The sample space are the 12 cans. So, $N(S_1)=12$ There are 3 cans filled with diet soda. Now, consider the event "can 1 contains diet soda". $N(can~1~contains~diet~soda)=3$ Using the Classical Method (page 259): $P(can~1~contains~diet~soda)=\frac{N(can~1~contains~diet~soda)}{N(S_1)}=\frac{3}{12}=\frac{1}{4}$ - Second can: The sample space are the 11 remaing cans. So, $N(S_2)=11$ There are 2 remaining cans filled with diet soda. Now, consider the event "can 2 contains diet soda". $N(can~2~contains~diet~soda~|~can~1~contains~diet~soda)=2$ Using the Classical Method (page 259): $P(can~2~contains~diet~soda~|~can~1~contains~diet~soda)=\frac{N(can~2~contains~diet~soda~|~can~1~contains~diet~soda)}{N(S_2)}=\frac{2}{11}$ Now, using the General Multiplication Rule (page 289): $P(both~cans~contain~diet~soda)=P(can~1~contains~diet~soda)\times P(can~2~contains~diet~soda~|~can~1~contains~diet~soda)=\frac{1}{4}\times\frac{2}{11}=\frac{2}{44}=\frac{1}{22}\approx0.04545$