Answer:
P(both soccer and basketball) = [tex]\frac{18}{31}[/tex]
Step-by-step explanation:
Let B, S denote number of students who play basketball and soccer.
As 31 play basketball, 59 play soccer and 18 of the students play both basketball and soccer,
[tex]n(B)=21\\n(S)=59[/tex]
n(B∩S) = 18
To find P (Soccer |Basketball) that is P(S∩B),
use P(S∩B) = P(both soccer and basketball)/ P(B)
P(both soccer and basketball) = Number of students who play both soccer and basketball / Total number of students
= [tex]\frac{18}{145}[/tex]
Also,
P(B) = Number of students who play basketball / Total number of students
= [tex]\frac{31}{145}[/tex]
So,
P(S∩B) = [tex]\frac{\frac{18}{145} }{\frac{31}{145} }=\frac{18}{31}[/tex]
That is
P(both soccer and basketball) = [tex]\frac{18}{31}[/tex]
