Answer
$P(both~cans~contain~regular~soda)=\frac{6}{11}\approx0.54545$
Work Step by Step
- First can:
The sample space are the 12 cans. So, $N(S_1)=12$
There are 9 cans filled with regular soda.
Now, consider the event "can 1 contains regular soda". $N(can~1~contains~regular~soda)=9$
Using the Classical Method (page 259):
$P(can~1~contains~regular~soda)=\frac{N(can~1~contains~regular~soda)}{N(S_1)}=\frac{9}{12}=\frac{3}{4}$
- Second can: The sample space are the 11 remaing cans. So, $N(S_2)=11$
There are 8 remaining cans filled with regular soda.
Now, consider the event "can 2 contains regular soda". $N(can~2~contains~regular~soda~|~can~1~contains~regular~soda)=8$
Using the Classical Method (page 259):
$P(can~2~contains~regular~soda~|~can~1~contains~regular~soda)=\frac{N(can~2~contains~regular~soda~|~can~1~contains~regular~soda)}{N(S_2)}=\frac{8}{11}$
Now, using the General Multiplication Rule (page 289):
$P(both~cans~contain~regular~soda)=P(can~1~contains~regular~soda)\times P(can~2~contains~regular~soda~|~can~1~contains~regular~soda)=\frac{3}{4}\times\frac{8}{11}=\frac{24}{44}=\frac{6}{11}\approx0.54545$
