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Suppose that among the 5000 students at a high school, 1200 are taking an online class and1700 prefer watching basketball to watching football. Taking an online class and preferringbasketball are independent. How many students are taking an online course and preferbasketball to football?

Sagot :

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

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