Answer
$P(both~televisions~work)=0.4$
$P(at~least~one~TV~does~not~work) = 0.6$
Work Step by Step
The sample space = {TV 1, TV 2, TV 3, TV 4, TV 5, TV 6}. So, N(S) = 6
Two televisions are defective. So, N(defective) = 2 and N(work) = 6 - 2 = 4.
Now, a television is randomly selected:
$P(1st~work)=\frac{N(work)}{N(S)}=\frac{4}{6}=\frac{2}{3}$.
Now, there are five remaining televisions.
The new sample space = {TV 1, TV 2, TV 3, TV 4, TV 5}. So, N(S) = 5. Two of them are defective. So, N(defective) = 2 and N(work) = 5 - 2 = 3.
Now, a second television is randomly selected:
$P(2nd~work~|~1st~work)=\frac{N(work)}{N(S)}=\frac{3}{5}$.
Using the General Multiplication Rule (see page 289):
$P(both~televisions~work)=P(1st~work)\times P(2nd~work~|~1st~work)=\frac{2}{3}\times\frac{3}{5}=\frac{2}{5}=0.40$.
The event "both televisions work" is the complement of "at least one does not work".
Now, using the Complement Rule (see page 275):
$P(at~least~one~does~not~work)=1 - P(both~televisions~work)=1-0.4=0.6.$
