Question

At Monroe High School, 62% of all students participate in after school sports and 11% participate in both after school sports and student council. What is the probability that a student participates in student council given that the student participates in after school sports?

Answer 1

Answer with Step-by-step explanation:

At Monroe High School, 62% of all students participate in after school sports and 11% participate in both after school sports and student council.

A: students participate in after school sports

P(A)=0.62

B:students participate in student council

A∩B:students participate in both after school sports and student council.

P(A∩B)=0.11

B/A: student participates in student council given that the student participates in after school sports

baye's theorem says that

P(A∩B)=P(B/A)×P(A)

0.11=P(B/A)×0.62

⇒ P(B/A)=0.18

Hence, probability that student participates in student council given that the student participates in after school sports is:

0.18

Answer 2

Hey there!

Before you start solving anything, you need to identify which situation you want to call event A and which you want to call event B. I usually just do it in the order of the events as they're given to me in the question, so:

A = Student participates in student council
B = Student participates in after school sports

Any problems that contain the word "given" in the question portion will want you to refer to P(A | B) = P(A ∩ B)/P(B). P(A | B) literally means "probability of event A, given that event B has occurred." P(A ∩ B) is the probability of event A and B happening, and P(B) is just the probability of event B happening. We've been given all of that, so:

P(A | B) = P(A ∩ B)/P(B)
P(A | B) = 11% / 62%
P(A | B) = 0.11 / 0.62
P(A | B) = 0.18

There will be about an 18% chance that a student participates in student council, given that the student participates in after school sports.

Hope this helped you out! :-)