Answer:
C. Playing soccer and basketball are not independent since P(soccer|basketball) ≠ P(soccer) and P(basketball|soccer) ≠ P(basketball) .
Step-by-step explanation:
We are given the data of athletic people that played soccer and basketball.
Let,
P(A) = probability of people who played soccer
P(B) = probability of people who played basketball
P(A∩B) = probability of people who played both soccer and basketball.
Now, we find the probability of soccer | basketball
i.e. [tex]P(A|B)=\frac{P(A \bigcap B)}{P(B)}[/tex]
i.e. [tex]P(A|B)=\frac{0.3}{0.7}[/tex] = i.e. [tex]P(A|B)=0.43[/tex]
As, [tex]P(A|B)=0.43[/tex] ≠ [tex]P(B)=0.7[/tex]
So, options A, B and D are discarded.
Moreover, i.e. [tex]P(B|A)=\frac{P(B \bigcap A)}{P(A)}[/tex]
i.e. [tex]P(B|A)=\frac{0.3}{0.5}[/tex] = i.e. [tex]P(B|A)=0.375[/tex]
i.e. [tex]P(B|A)=0.375[/tex] ≠ [tex]P(A)=0.5[/tex]
Hence, playing soccer and basketball are not independent.
