# Chapter 5 - Section 5.4 - Assess Your Understanding - Applying the Concepts - Page 294: 28b

$P(both~cans~contain~regular~soda)=\frac{6}{11}\approx0.54545$

#### 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): $P(both~cans~contain~regular~soda)=P(can~1~contains~regular~soda)\times P(can~2~contains~regular~soda~|~can~1~contains~regular~soda)=\frac{3}{4}\times\frac{8}{11}=\frac{24}{44}=\frac{6}{11}\approx0.54545$

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