What is the probability that I have watched neither news? I watch the 6o'clock news $2/3$ of the time, the 11o'clock news $1/2$ of the time, and the [$6$ and $11$ news] $1/3$ of the time. What is $P(\text{neither news})$ * * * What is the probability that I watch neither news? This problem can be tackled two ways (one is correct, but I'm not sure which one). Originally I thought that $P(\text{neither news})=1-P(\text{both news})=1-1/3=2/3$ But then I think that $P(\text{neither news})=P(\text{don't watch 6})\cdot P(\text{don't watch 11})=(1-2/3)\cdot (1-1/2)=1/6$ Which one is correct here, and why? Is the opposite event of **neither** considered to be both? In that case first one is correct. If this is not the case, second one is correct.

You first working is wrong because you have identified the wrong complement.

The second working is correct but it is unclear to me whether you are aware of the reasoning. It turns out the event of watching each news is independent in this case.

Alternatively, not watching both news is the complement of watching at least one news.

$$P( A' \cap B')=1-P(A \cup B) = 1-P(A)-P(B)+P(A \cap B)$$