Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

In a recent survey in a Statistics class, it was determined that only 70% of the students attend class on Fridays. From past data it was noted that 88% of those who went to class on Fridays pass the course, while only 20% of those who did not go to class on Fridays passed the course. a. What percentage 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? (Hint: the above problem is based on conditional probability concept and the Bayes theorem)

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

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.