Let [tex]V[/tex] be the set of people infected with the virus, and [tex]T[/tex] the set of people who get a positive result (true or false) when tested for the virus. Any person tested for the virus will get one or the other result and not both, so the events [tex]V\cap T[/tex] and [tex]V\cap T^C[/tex] are mutually exclusive:
[tex]V=(V\cap T)\union(V\cap T^C)\implies |V|=|V\cap T|+|V\cap T^C|[/tex]
where [tex]|\cdot|[/tex] denotes the size/cardinality of a set.
We're told that [tex]|V|=50[/tex] and [tex]|V\cap T|=48[/tex], which means [tex]|V\cap T^C|=50-48=2[/tex].
Similarly, we have
[tex]V^C=(V^C\cap T)\union(V^C\cap T^C)\implies|V^C|=|V^C\cap T|+|V^C\cap T^C|[/tex]
and we know [tex]|V^C|=200-50=150[/tex] and [tex]|V^C\cap T^C|=135[/tex], so [tex]|V^C\cap T|=150-135=15[/tex].
The probability of each possible event is obtained by dividing the size of each corresponding set by the total 200. So we have the following table:
[tex]\begin{matrix}&V&V^C&\text{total}\\T&48&15&63\\T^C&2&135&137\\\text{total}&50&150&200\end{bmatrix}[/tex]
The events [tex]V[/tex] and [tex]T[/tex] are independent if [tex]P(V\cap T)=P(V)\cdot P(T)[/tex]. From the table above, we find that
[tex]P(V\cap T)=\dfrac{48}{200}[/tex]
and
[tex]P(V)\cdot P(T)=\dfrac{50}{200}\cdot\dfrac{63}{200}=\dfrac{63}{800}[/tex]
which are not equal, so the events are not independent.
