Answer
$P(both~cans~contain~diet~soda)=\frac{1}{22}\approx0.04545$
Work Step by Step
- 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):
$P(both~cans~contain~diet~soda)=P(can~1~contains~diet~soda)\times P(can~2~contains~diet~soda~|~can~1~contains~diet~soda)=\frac{1}{4}\times\frac{2}{11}=\frac{2}{44}=\frac{1}{22}\approx0.04545$
