Solution:
Given:
[tex]\begin{gathered} total=5000 \\ n_{online\text{ class}}=1200 \\ n_{basketball}=1700 \end{gathered}[/tex]Since both events are independent, then;
[tex]P(A\cap B)=P(A)\times P(B)[/tex]Hence,
[tex]\begin{gathered} P(online)=\frac{1200}{5000} \\ P(basketball)=\frac{1700}{5000} \\ Hence,\text{ } \\ P(online\cap basketball)=\frac{1200}{5000}\times\frac{1700}{5000} \\ P(online\cap basketball)=\frac{51}{625} \\ P(online\cap basketball)=0.0816 \end{gathered}[/tex]The number of students taking an online course and prefer basketball to football is;
[tex]\begin{gathered} n(online\cap basketball)=0.0816\times5000 \\ n(online\cap basketball)=408 \end{gathered}[/tex]Therefore, the number of students taking an online course and prefer basketball to football is 408
