1. Mar 28, 2010

### hpt001

(Moderator's note: thread moved from "Set Theory, Logic, Probability, Statistics ")

"A lecturer gives spot quizzes in randomly selected classes. In those quizzes he asks questions from the material covered in the previous class. From experience it is known that a student has 0.85 chance of passing the quiz if the student was present in the last class. For a student who was absent last class, that chance is only 0.20. There are 30 students in the class today. Five of those students were absent in the last class. The lecturer has given a spot quiz today."

What is the probability that more than 25 students will pass?

Last edited by a moderator: Mar 28, 2010
2. Mar 28, 2010

### CompuChip

Can you calculate the probability that of the 30 students, 3 that skipped class fail (the other two pass) and 2 that did attend also fail (the other 23 pass)?

Can you do the same, more generally, in the situation where n of the students who missed class and a of the students who attended will fail the test (so 5 - n and 25 - a pass)?

3. Mar 29, 2010

### hpt001

This is the way I tried to solve this.
Let
X = The number of students that pass out of 25 students present last class.
Y = The number of students that pass out of 5 students absent last class.
T = Total number of students pass out of 30 students.
X~Bin(25,0.85) and Y~Bin(5,0.20)

PX(x) = 25Cx(0.85x)(0.1525-x ), x = 0,1,2,......25

PY(y) = 5Cy(0.20y)(0.805-y ), y = 0,1,2,3,4,5

The # of passes can be any of following
(Present Absent) Last class
25 and 1 or 2 or 3 or 4 or 5
24 and 2 or 3 or 4 or 5
23 and 3 or 4 or 5
22 and 4 or 5
21 and 5

Pr(T>25) = Pr[(X=25)$$\cap$$(Y=1)$$\cup$$(X=25)$$\cap$$(Y=2)$$\cup$$(X=25)$$\cap$$(Y=3)...............................$$\cup$$(X=25)$$\cap$$(Y=5)]

= Pr(X=25).Pr(Y=1)+Pr(X=25).Pr(Y=2)+.......................+Pr(X=22).Pr(Y=5)+Pr(X=21).Pr(Y=5)

But I'm not sure that whether this method is correct or not.