Question

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)

Answer 1

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ 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

$p = 0.88*0.7 + 0.2*0.3 = 0.676$

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 $P(A) = 0.676$

Probability that passed and attended classes on Friday.

88% of 70%

This means that:

$P(A \cap B) = 0.88*0.7 = 0.616$

Then

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.616}{0.676} = 0.9112$

0.9112 = 91.12% probability that he/she attended classes on Fridays