The number of surveyed teachers who play videogames is 6, according to the table; therefore,
[tex]P(Teacher\cap Videogames)=\frac{6}{150}=\frac{1}{25}=0.04=4\%[/tex]
On the other hand, if A and B are two independent events,
[tex]\begin{gathered} A,B\rightarrow\text{ independent events} \\ \Rightarrow P(A\cap B)=P(A)P(B) \end{gathered}[/tex]
Thus, in our case,
[tex]\begin{gathered} P(Teacher)P(Videogames)=\frac{6+14}{150}*\frac{48+45+6}{150}=\frac{20}{150}*\frac{99}{150}=\frac{11}{125}=0.088 \\ \Rightarrow P(Teacher\cap Videogames)\ne P(Teacher)P(Videogames) \end{gathered}[/tex]
Therefore, the two events are not independent.
Hence, the answer is the first option, 4%, and not independent.
