Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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
[tex]p = 0.88*0.7 + 0.2*0.3 = 0.676[/tex]
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 [tex]P(A) = 0.676[/tex]
Probability that passed and attended classes on Friday.
88% of 70%
This means that:
[tex]P(A \cap B) = 0.88*0.7 = 0.616[/tex]
Then
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.616}{0.676} = 0.9112[/tex]
0.9112 = 91.12% probability that he/she attended classes on Fridays
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
