Question

4. One flaw in attendance mechanisms is that UC Merced students work jobs to make wages. Suppose that there is a student whose job has a time conflict with our course, and it is tough to predict when we will see that student the next time. So far this semester, that student has a 20.59% attendance score. What is the probability that we will observe 3 absences and then observe the student in class on the fourth lecture. 5. According to the New York Times, eleven percent of parents helped their college children write an essay. If I collect seven essays, that is the probability that at most two of those submissions had parental assistance?! 6. Suppose that at a famous aquarium, a docent found that 35 percent of elementary school children know the difference between dolphins and porpoises. If a docent has a tour group that has 9 elementary school children, what is the probability that at least two of the children know the difference between dolphins and porpoises? 7. Even at conferences full of physicians, female doctors noticed that they were introduced as “Doctor” about 49% of the time.2 Visualize a female physician networking at such an event. What is the probability that she will be introduced as “Doctor” 3 times before not being addressed by the title on the fourth introduction.

Answer 1

There are many different questions you have posted so I shall answer the first one and if you need the assistance with all of them then please post them seperately.

The student's attendance score tells us the probability of that student attending the class on any day.

i.e. Probability of student being present in class on any day, say P(P) = 20.59% = 0.2059

So, Probability of student being Absent in class on any day, say P(A) = 1-P(P) = 1-0.2059= 0.7941

If we assume the event of student being present/absent in a class on any day is INDEPENDENT to (s)he being present/absent in a class on previous days then

Probability of student being absent in class on 2 consecutive days = P(A1 ∩ A2) = P(A1)*P(A2)

Similarly, Probability of student being absent in class on 3 consecutive days = P(A1 ∩ A2 ∩ A3) = P(A1)*P(A2)*P(A3)

Similarly, Probability of student being absent in class on first 3 consecutive days and being present on 4th day = P(A1 ∩ A2 ∩ A3 ∩ P4) = P(A1)*P(A2)*P(A3)*P(P4) = 0.2059 * 0.2059 * 0.2059 * 0.7941 = 0.0069 = .69% (i.e. less than 1% chance)