The final exam grade cutoffs were:

    grade cutoff
    A     130
    AB    118
    B     106
    BC    96
    C     80
    D     65
  

As announced in lecture today, the problems below will not be on Exam 2 (or the final), since they are too time consuming for an exam situation.

  section  problems
   3.2      34, 35
   3.3      31,32
   3.4      32,33,34
   3.5      25,34
   4.1      33

Section 3.3 problem 31 states:

  A reduced deck of 36 cards consists of all numbered cards,
  that is, four cards numbered 2, four cards numbered 3, ...,
  four cards numbered 10.  Five cards are dealt, one after another
  without replacement, and the number on each card is noted.
  Find the probability that the numbers of the cards dealt
  are strictly increasing.  That is, each card dealt has a larger
  number than the previous card.

Solution.

  Let A = event that hand is strictly increasing
  Let B = event that hand could be reordered so that
          it is strictly increasing.

  Pr[A] = Pr[A and B]  (since A implies B)
        = Pr[A|B] Pr[B].

  Pr[B] = (number of hands that consist of different numbers) divided by
          (number of hands)

  (number of hands) = C(36,5).

  (number of hands that consist of different numbers)
   = C(9,5) (i.e. the number of choices of 5 digits from 9 possible digits)
     times 4*4*4*4*4 (i.e. the number of choices of suite for the 5 digits)

  So Pr[B] = (C(9,5) * 4^5) divided by
             C(36,5)

  Pr[A|B] = 1/(5!).

  So Pr[A] = (C(9/5) * 4^5) divided by
             (5! * C(36,5))
           = 9*8*7*6*5 * 4^5 divided by
             36*35*34*33*32 * 5!

Section 2.3 problem 31 states:

  The faculty at GSU is divided into four schools:
  Humanities, Science, Music, and Athletics.
  The faculty council has 8 members, and it must
  contain at least 1 faculty member from each school.
  In how many different ways (in terms of school
  representatives) can the council be constituted?
  (For example, one possible way is to have 2 members
  from each of the 4 schools.)

Answer: C(7,3)

Solution:

  For the purposes of the problem the makeup of a council
  is the number of representatives from each school.
  Make the definitions:

    H = number of representatives from humanities
    S = number of representatives from sciences
    M = number of representatives from music
    A = number of representatives from athletics

  The problem says that we must find the number of ways to
  choose H, S, M, and A subject to the requirements that

    H+S+M+A = 8,
    H ≥ 1,
    S ≥ 1,
    M ≥ 1,
    A ≥ 1,

  That is, we need to count the number of ways that
  four positive numbers can add to 8.
  Here's a trick to count up the number of ways:

  Put down a row of 8 chairs where we will seat the council members.
  We will seat the people from each school next to each other like this:

    h h s s s m m a

  We need to count the number of ways of dividing the chairs
  into 4 groups of adjacent chairs.
  We do so by placing 3 dividers in the 7 slots between the chairs,
  like this:

    h h|s s s|m m|a

  There are C(7,3) ways to place the dividers.

Section 2.4 problem 33 states:

  There are 5 unique positions available within a Fortune 500
  company.  Two of these positions must be filled with women,
  and the others could be filed with either men or women.
  Six men, including Shawn, have applied, and 7 women
  have applied.

  (a) In how many ways can the positions be filled?

  (b) If the positions are filled at random,
      what is the probability that Shawn is appointed?

Looking at the answer, it appears that the requirement is that
two *specific* positions must be filled with women (so it is not
enough to ensure that at least two of the positions are filled
with women).

Solution to part a:

  Let S = set of allowed outcomes ("sample space").
  The outcomes in S are uniquely given by two successive experiments:
  
    stage  # outcomes   action
    =====  ==========   ==============================================
    S1     P(7,2)       select two women for the famale-only positions
    S2     P(11,3)      select three of the remaining applicants
                          for the remaining positions.
  
  So the answer is n(S) = P(7,2) * P(11,3) = (7*6)*(11*10*9).

Solution to part b:

  Let E = set of outcomes that involve Shawn getting picked (the "event").
  All outcomes in E are uniquely given by three successive experiments:

    stage  # outcomes   action
    =====  ==========   ==============================================
    S1     P(7,2)       select two women for the famale-only positions
    S2     3            select a position for Shawn
    S2     P(10,2)      select two of the remaining applicants
                          for the remaining positions.
  
  So n(E) = P(7,2)*3*P(10,2) = (7*6)*3*(10*9)

  Assuming that all outcomes in the sample space S are equally
  likely, the probability that Shawn is picked is thus

  n(E)/n(S) = 3/11.