The number of surveyed teachers who play videogames is 6, according to the table; therefore,
![P(Teacher\cap Videogames)=(6)/(150)=(1)/(25)=0.04=4\%](https://img.qammunity.org/2023/formulas/mathematics/college/55qj2e8t37ejfssn98me2rycbf8nps24ur.png)
On the other hand, if A and B are two independent events,
![\begin{gathered} A,B\rightarrow\text{ independent events} \\ \Rightarrow P(A\cap B)=P(A)P(B) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w8u7aljd9o1pth3gr79z8r0gyke767q4e2.png)
Thus, in our case,
![\begin{gathered} P(Teacher)P(Videogames)=(6+14)/(150)*(48+45+6)/(150)=(20)/(150)*(99)/(150)=(11)/(125)=0.088 \\ \Rightarrow P(Teacher\cap Videogames)\\e P(Teacher)P(Videogames) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2qpsfobfnqjqj1ecz13nrpe4m02jatbpf4.png)
Therefore, the two events are not independent.
Hence, the answer is the first option, 4%, and not independent.