In a recent survey in a Statistics class, it was determined that only 60% of the students attend class on Fridays. From past data it was noted that 98% 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?

To calculate the answers to these questions, we can use conditional probability.

a. To find the percentage of students expected to pass the course, we need to calculate the overall probability of passing the course. This can be calculated using the law of total probability.

Let's assume there are 100 students in the class.

Out of these, only 60% attend class on Fridays, so there are 60 students who attend class on Fridays.

Among those who attend class on Fridays, 98% pass the course. So, out of the 60 students who attend class on Fridays, 98% of them pass, which is 0.98 * 60 = 58.8 students.

The remaining 40% of students, who do not attend class on Fridays, are 40 students. Among these students, only 20% pass the course. So, out of the 40 students who don't attend class on Fridays, 20% of them pass, which is 0.20 * 40 = 8 students.

Therefore, the total number of students expected to pass the course is 58.8 + 8 = 66.8 students.

To find the percentage of students expected to pass the course, we divide the number of students expected to pass (66.8) by the total number of students (100) and multiply by 100:

(66.8 / 100) * 100 = 66.8% of students are expected to pass the course.

So, the percentage of students expected to pass the course is 66.8%.

b. To find the probability that a student attended classes on Fridays given that they passed the course, we can use Bayes' theorem.

Let's denote:
P(F) as the probability of attending class on Fridays.
P(P) as the probability of passing the course.

We know:
P(F) = 0.6 (60% of students attend class on Fridays)
P(P|F) = 0.98 (given that a student attended class on Fridays, the probability of passing)
P(P|not F) = 0.2 (given that a student did not attend class on Fridays, the probability of passing)

We want to find:
P(F|P) (the probability of attending class on Fridays given that a student passed the course).

Using Bayes' theorem:
P(F|P) = (P(F) * P(P|F)) / (P(F) * P(P|F) + P(not F) * P(P|not F))

P(not F) = 1 - P(F) = 1 - 0.6 = 0.4 (40% of students do not attend class on Fridays)

Substituting the values:
P(F|P) = (0.6 * 0.98) / (0.6 * 0.98 + 0.4 * 0.2)
= 0.588 / (0.588 + 0.08)
= 0.588 / 0.668
≈ 0.8794

So, the probability that a student attended classes on Fridays given that they passed the course is approximately 0.8794, or about 87.94%.

a. To calculate the percentage of students expected to pass the course, we need to consider both the percentage of students who attend class on Fridays and the pass rate for those who attend class on Fridays, as well as the percentage of students who do not attend class on Fridays and the pass rate for those who do not attend class on Fridays.

Let's denote:
P(F) = Percentage of students attending class on Fridays = 60% = 0.6
P(P | F) = Pass rate for students attending class on Fridays = 98% = 0.98
P(not F) = Percentage of students not attending class on Fridays = 100% - P(F) = 40% = 0.4
P(P | not F) = Pass rate for students not attending class on Fridays = 20% = 0.2

Now we can calculate the expected percentage of students who pass the course using the law of total probability:
P(P) = P(P | F) * P(F) + P(P | not F) * P(not F)
= 0.98 * 0.6 + 0.2 * 0.4
= 0.588 + 0.08
= 0.668

Therefore, approximately 66.8% of students are expected to pass the course.

b. To calculate the probability that a person attended classes on Fridays given that they passed the course, we can use Bayes' theorem:

P(F | P) = (P(P | F) * P(F)) / P(P)
= (0.98 * 0.6) / 0.668
= 0.588 / 0.668
≈ 0.88

Therefore, given that a person has passed the course, the probability that he/she attended classes on Fridays is approximately 88%.

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