Respuesta :
Answer:
71% probability that the athlete is either a football player or a basketball player
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that an athlete is a football player.
B is the probability that an athlete is a basketball players.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that an athlete plays football but not basketball and [tex]A \cap B[/tex] is the probability that an athlete plays both these sports.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
23% of the athletes play both football and basketball.
This means that [tex]A \cap B = 0.23[/tex]
26% are basketball players
This means that [tex]B = 0.26[/tex]. So
[tex]B = b + (A \cap B)[/tex]
[tex]0.26 = b + 0.23[/tex]
[tex]b = 0.03[/tex]
68% of the athletes are football players
This means that [tex]A = 0.68[/tex]. So
[tex]A = a + (A \cap B)[/tex]
[tex]0.68 = a + 0.23[/tex]
[tex]a = 0.45[/tex]
What is the probability that the athlete is either a football player or a basketball player
[tex]A \cup B = a + b + (A \cap B)[/tex]
[tex]A \cup B = 0.45 + 0.03 + 0.23 = 0.71[/tex]
71% probability that the athlete is either a football player or a basketball player
Answer:
71%
Step-by-step explanation:
Applying Venn's diagram
probabilty that an athlete plays either football or basketball
P ( A ∪ B ) = a + b + ( A ∩ B ) equation 1
note:
P ( A ) = a + ( A ∩ B )
( A ∩ B ) = 23% = 0.23
a = P( A ) - 0.23 = 0.68 - 0.23 = 0.45
P ( B ) = b + ( A ∩ B )
b = P ( B ) - ( A ∩ B ) = 0.26 - 0.23 = 0.03
back to equation 1
P ( A ∪ B ) = 0.45 + 0.03 +( 0.23 )
= 0.48 + ( 0.23 ) = 0.71 = 71%
P( A ) = probability of football players
P ( B ) = probability of basketball players
a = probability of playing football but not basketball
b = probabilty of playing basketball but not football
