Answer
$P(exactly~two~are~Merlot)=\frac{7}{22}\approx0.3182$
Work Step by Step
The order in which the bottles are selected does not matter and no bottle can be selected more than once.
Now we have 2 Merlot and 1 Cabernet.
The number of combinations of 5 distinct bottles of Merlot taken 2 at a time:
$_5C_2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!\times3!}=\frac{5\times4\times3\times2\times1}{2\times6}=10$
The number of combinations of 7 distinct bottles of Cabernet taken 1 at a time:
$_7C_1=\frac{7!}{1!(7-1)!}=\frac{7!}{1!\times6!}=\frac{7\times6\times5\times4\times3\times2\times1}{1\times6\times5\times4\times3\times2\times1}=7$
Using the Multiplication Rule of Counting (page 298):
$N(two~are~Merlot~and~one~is~Cabernet)=10\times7=70$
The number of combinations of 12 distinct bottles (5 Merlot and 7 Cabernet) taken 3 at a time:
$N(S)=~_{12}C_3=\frac{12!}{3!(12-3)!}=\frac{12!}{3!\times9!}=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times9\times8\times7\times6\times5\times4\times3\times2\times1}=\frac{12\times11\times10}{3\times2\times1}=220$
Using the Classical Method (page 259):
$P(exactly~two~are~Merlot)=P(two~are~Merlot~and~one~is~Cabernet)=\frac{N(two~are~Merlot~and~one~is~Cabernet)}{N(S)}=\frac{70}{220}=\frac{7}{22}\approx0.3182$
