Solutions 3

7.1:

19: First find the number of hands with 5 different kinds: we choose 5
    kinds out of the possible 13, and then each kind can be any of 4
    suits. So there are (13 choose 5) * 4^5 such hands.
    In class we found how many flushes there are: we choose 5 kinds
    out of 13 and then the first card can be any of 4 suits. Once that
    is chosen, there are no further choice. So there are (13 choose 5)
    * 4 flushes. 
    To make a straight, we have to choose where it starts, which
    can be any kard A,2,...,10. That makes 10 ways. Each card can be
    any of 4 suits, so we get 10*4^5.
    We also need to know the number of straight flushes, which as you
    find easily is 10*4. 
    Inclusion/exclusion then tell us that the number of handsd without
    a pair, a flush or a straight is 
    (13 choose 5) * 4^5  - (13 choose 5) * 4 - 10*4^5 + 10*4.
    Divide this by (52 choose 5).

32: There are (100 choose 3) ways of picking 3 winners. There is one
    way to pick the three people Kumar, Janice and Pedro. So the
    probability if 1/(100 choose 3).


14: There are 13 choose 5 ways of picking 5 different kinds of
    cards. Each of them can come from 4 different suits, so the number of
    getting 5 cards of different kinds is (13 choose 5)*4^5. There are 52
    choose 5 hands altogether, so the probability is 
    (13 choose 5)*4^5/(52 choose 5).

28: Suppose the commission has chosen secretely 11 numbers. Then we go
    and pick 7 of the 80, allowing for 80 choose 7 possibilities. The
    cases in which we win are those where the 7 come from the 11
    preselected ones. There are 11 choose 7 such good cases. So the
    probability is (11 choose 7) / (80 choose 7).

30: Suppose the computer has chosen his 6 numbers. We go and pick in
    one of (40 choose 6) ways our 6. Then what are the good cases?
    When 5 of our 6 are in the computer's 6, and one of our 6 is
    outside the computer's 6. That is, we have 6 choose 5 ways to get
    the 5 correct ones, and 34 choose 1 way to get the wrong one. 
    So the probability is (6 choose 5)*(34 choose 1)/(40 choose 6).

36: There are 5 ways out of 6^2 to roll an 8 with 2 dice. There are 21
    ways out of 6^3 to roll an 8 with 3 dice. (Just write them out...)
    Hence the two probabilities are 21/216 and 5/36. The bigger is
    5/36.

38: a) E1 has 4 good cases, E2 has 4 good cases, their intersection
    has 2 good cases. So p(E1)=4/8, p(E2)=4/8, p(E1 intersect
    E2)=2/8. The equality for independence olds since 4/8 * 4/8 =2/8.
    b) p(E1)=4/8, p(E2)=2/8, p(E1 intersect E2)=1/8. Again, the
    equality holds.
    c) p(E1)=4/8, p(E2)=2/8, p(E1 and E2)=0. The inequality fails,
    E1 and E2 are not independent.


8.5:

14: To find all words without "fish", "bird" and "rat" is equivalent
    to finding all permutations that contain one of the three words,
    because they will add up to 26!, the number of all permutations. 
    In other words, we want to find the number of elements in the
    union A1 union A2 union A3 where A1 are those with "fish", A2
    those with "rat", A3 those with "bird". 

    Let's count how many permutations contain "fish". This is like
    having 22 + 1 letters: the 22 that are not in "fish", and the
    letter "fish". So there are 23! words with fish. Similarly, there
    are 24! with "rat" and 23! with "bird". 
    Now let's count the permutations in A1 intersect A2 -- they are
    the ones with "fish" and "rat". In this case we have 26-7 letters
    plus the "letters" "fish" and "rat". So there are 21! permutations
    with "fish" and "rat". 
    To count A1 intersect A3 we note that one cannot have "fish" and
    "bird" in the same permutation since either "i" is followed by "s"
    or by "r". So there are no elements in that intersection.
    Similarly there are no permutations that contain "rat" and
    "bird". 
    Also, there are no permutations with all three words.
    It follows that the number of permutations that contain at least
    one of the three words is
    23! + 24! + 23! - 21! - 0 - 0 + 0
    following the principle
	      * add the numbers of elements in A1, A2, A3
	      * subtract the numbers of elements in pairwise
		intersections,
	      * add the number in triple intersection.
    To get the words that DON'T contain any of "fish", "bird", "rat",
    take the number found above and subtract from 26!.

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