Question

at a particular school with 200 male students, 58 play football, 40 play basketball, and 8 play both. what is the probability that a randomly selected male student plays neither sport?

Answer 1

this helps  (no play) = 121/200 = 0.6050

ifti doesn't I tried

Answer 2

Answer:

$\frac{11}{20}$

Step-by-step explanation:

Total male students = 200

Students who play football (F)= 58

Students who play basketball (B)= 40

Students who play both (F∩B)= 8

The number of students who play either sport is

$(F\cup B)=F+B-(F\cap B)$

$(F\cup B)=58+40-8=90$

The number of students who play neither sport is

$200-(F\cup B)=200-90=110$

The probability that a randomly selected male student plays neither sport is

$Probability=\dfrac{\text{Number of students who play neither sport}}{\text{Total male students}}$

$Probability=\dfrac{110}{200}$

$Probability=\dfrac{11}{20}$

Hence, the required probability is $\frac{11}{20}$.