Answer:
84%
Step-by-step explanation:
The probability that the selected player is a football player, P(F)=19%
The probability that the selected player is a basketball player, P(B)=79%
The probability that the selected player play both football and basketball,
[tex]P(B \cap F)=14\%[/tex]
We want to determine the probability that a randomly chosen athlete is either a football player or a basketball player, [tex]P(B \cup F)[/tex]
In probability theory
[tex]P(B \cup F)=P(B)+P(F)-P(B \cap F)\\=79\%+19\%-14\%\\=84\%[/tex]
The probability that a randomly chosen athlete is either a football player or a basketball player is 84%.