Answer
$P(no~Merlot)=\frac{7}{44}\approx0.1591$
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 no Merlot and 3 Cabernet.
The number of combinations of 7 distinct bottles of Cabernet taken 3 at a time:
$_7C_3=\frac{7!}{3!(7-3)!}=\frac{7!}{3!\times4!}=\frac{7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times4\times3\times2\times1}=35$
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):
$P(none~is~a~Merlot)=P(3~Cabernet)=\frac{N(3~Cabernet)}{N(S)}=\frac{35}{220}=\frac{7}{44}\approx0.1591$
