Question

Out of a group of 145 students that were surveyed about sports, 31 said they play basketball and 59 said they play soccer. 18 of the students who said they play basketball said they also play soccer. If a student is chosen at random, find the probability. P (Soccer |Basketball) P =

Answer 1

Answer:

P(both soccer and basketball) = $\frac{18}{31}$

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,

$n(B)=21\\n(S)=59$

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

= $\frac{18}{145}$

Also,

P(B) = Number of students who play basketball / Total number of students

= $\frac{31}{145}$

So,

P(S∩B) = $\frac{\frac{18}{145} }{\frac{31}{145} }=\frac{18}{31}$

That is

P(both soccer and basketball) = $\frac{18}{31}$