A college professor hands out a list of 10 questions, 5 of which will appear on

Question

A college professor hands out a list of 10 questions, 5 of which will appear on the final exam for the course. One of the students taking the course is pressed for time and can only prepare for 7 of the 10 questions Suppose that the professor chooses the 5 questions at random from the 10 (a) What is the probability that the student will be prepared for all 5 questions that appear on the final examination? than 3 questions that appear on the final examination? actly 4 questions that appear on the final examination? (b) What is the probability that the student will be prepared for fewer (c) What is the probability that the student will be prepared for ex-

Explanation / Answer

One of the students taking the course is pressed for time and can prepare for only 7 of the 10 questions on the list.

This is equivalent to saying divide the 10 questions into 7 prepared (P) + 3 unprepared (U).
Then the professor randomly chooses 5 from those 10 (without replacement, order irrelevant => permutation) .
(Think of choosing from a box containing 7 red balls and 3 blue balls, it's analogous.)

The number of permutations of choosing 5 questions (n prepared and (5-n) unprepared) from 7 prepared and 3 unprepared is:
W(n) = C(7,n) * C(3,5-n)
and obviously W(n) is only nonzero for 3n5
(W(1)=0=W(2), can't choose more than 3 unprepared)

The total number of permutations of choosing 5 questions from these 10 is W(3)+W(4)+W(5)
W(5) = C(7,5) * C(3,0) = 7*6/2! * 1 = 21
W(4) = C(7,4) * C(3,1) = 7*6*5*4/4! * 3 = 35*3 = 105
W(3) = C(7,3) * C(3,2) = 7*6*5/3! * 3*2/2! = 35*3 = 105
Thus total permutations = W(3)+W(4)+W(5) = 231

a) W(5)/(W(3)+W(4)+W(5)) = 21/231
b) 0, as commented above W(0)=W(1)=W(2)=0
c) W(4)/(W(3)+W(4)+W(5)) = 105/231

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