Answer:
a) 67.6% of students is expected to pass the course
b) 0.9112 = 91.12% probability that he/she attended classes on Fridays
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = \frac{P(A \cap B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
a. What percentage of students is expected to pass the course?
88% of 70%(attended class)
20% of 100 - 70 = 30%(did not attend class). So
![p = 0.88*0.7 + 0.2*0.3 = 0.676](https://tex.z-dn.net/?f=p%20%3D%200.88%2A0.7%20%2B%200.2%2A0.3%20%3D%200.676)
0.676*100% = 67.6%
67.6% of students is expected to pass the course.
b. Given that a person passes the course, what is the probability that he/she attended classes on Fridays?
Here, we use conditional probability:
Event A: Passed the course
Event B: Attended classes on Fridays.
67.6% of students is expected to pass the course.
This means that ![P(A) = 0.676](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.676)
Probability that passed and attended classes on Friday.
88% of 70%
This means that:
![P(A \cap B) = 0.88*0.7 = 0.616](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.88%2A0.7%20%3D%200.616)
Then
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.616}{0.676} = 0.9112](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.616%7D%7B0.676%7D%20%3D%200.9112)
0.9112 = 91.12% probability that he/she attended classes on Fridays