Solutions Assignment 4:

7.2:

2: If 3 is twice as likely as all other numbers,
   p(3)/2=p(1)=p(2)=p(4)=p(5)=p(6). Let x=p(1). 
   Since the sum of all p(i) is 1, x+x+2x+x+x+x+x=1, so x=1/7. 


10: a) Note that the 13th letter is "m", and there are 26! ways of
    arranging the 26 letters of the alphabet. 
    INTERPRETATION 1 of the question:
    The first 13 letters could be
    (a,...,m) or (b,...,n), or ..., or (n,...,z).
    These are 14 cases. In each case, the other 13 letters can be
    arranged in 13! ways, so the answer is 13!*14/26!.
    INTERPRETATION 2: What if the first 13 letters don't have to be
    consecutive? Then there are 26 choose ways of picking the first
    13, and for each such subset there is precisely one way of writing
    them in order. The second part has still 13!, and there are still
    13! together. So, (26 choose 13)*13!/26!. 
    b) There are 24 letters to be placed, so 24!/26!.
    c) Assume first that a comes before z. Then they could be in
    positions 1 and 2, or 2 and 3, or ..., or 25 and 26. These are 25
    choices, and for each the other 24 letters can be arranged in 24!
    ways. So we get 25! ways for a coming before z. So altogether
    the probability is 2*25!/26!.
    d) From c) we know there are 2*25! ways for a dn b not be
    neighbors. Hence there are (26!-2*25!) ways giving probability
    24*25!/26!.
    e) First suppose that a comes before z. Then there are possible
    positions (1,25), (1,26) and (2,26) for (a,z). There are 24! ways
    to place the other 24 letters, so 24!*3 ways. Counting also when z
    comes before a, the probability is 2*3*24!/26!.
    f) There are 26 choose 3 ways to choose the 3 places where a,b,z
    go. Once that is done, there 2 ways to place the 3 letters into
    the 3 slots, (z,a,b) and (z,b,a). In all cases, there are 23! ways
    of placing the other letters. So we get 2*(26 choose 3)*23!/26!. 


12: Since E union F contains the event E, its probability is at least
    as much as p(E)=0.8.
    Let G be E intersect F. IF p(G)<0.4, then surely p(E\G)>0.4, since
    if 80% of all people live in E but fewer than 40% belong to E and
    G then more than 40% live in E but not in G.
    If follows that the entire sample space can be divided into three
    regions: E\G, F, and H=outside both E and F. These regions are
    disjoint in the sense that every element of the sample space is in
    one and precisely one of the 3 regions. That means that
    p(S)=p(E\G)+p(F)+p(H).  
    But now p(S)=1 and p(E\G)>0.4 while p(F)=0.6. Hence E\G and F
    alone give a probability of more than 1, which is just not an
    option. So our assumption "IF" above must have been wrong and so
    p(G) is at least 0.4.
 
30: a) If no 0 shows, 10 trials have produced a 1. Each
    time this happens with probability 1/2, so p(no zero in 10 trials)=1/2^10.
    b) Now ten 1's have a probability of (0.6)^10. 
    c) No zero after 1 trial: 1/2.
       No zero after 2 trials: (1/2)*(1/4).
       No zero after 3 trials: (1/2)*(1/4)*(1/8)
       ...
       No zero after 10 trials: (1/2)*(1/4)*...*(1/2^10). 

18: a) There are 7 choices for a birthday for each of the 2
    people. That gives a total of 49 possibilities. Then out of these
    49 there are 7 good cases, "along the diagonal". So the
    probability is 7/49.
    b) Let's count as in the birthday problem. First off, if we have 8
    or more people in a room, then there will be 2 with the same
    birthday because there are only 7 days in the week. 
    If there are 0 or 1 persons in the room, they cannot have 2 with
    the same birthday. If there are 2 in the room, it's part a). If
    there are 3 in the room, we have (7 choose 3) waysof choosing
    three different weekdays, and then (7 choose 3) * 3! to assign the
    three days to the 3 people. Therefore there are 7^3 - (7
    choose 3)*3! cases where 2 people share a birthday. So the
    probability is (7^3-(7 choose 3)*3!) / 7^3.
    In fact, for all numbers 2,...,7 we see that the probability is 
    (7^k - (7 choose k)*k!) / 7^k. 
    c) Just plug in the numbers 2,3,4,5,6,7 into the answer of part b)
    and see when the result is at least 1/2. 

24: To There are 32 = 2^5 outcomes for flipping a coin 5 times. There
    are 16=1*2^4 cases where the first flip is a tail. There are 5
    ways of having exactly 4 heads. There is one way of having a tail
    first and still 4 heads. So the conditional probability of having
    4 heads when starting with a tail is (1/32) / (16/32).

26: An odd number of 1's occurs in the strings (1,0,0), (0,1,0),
    (0,0,1), (1,1,1). There are 2 of them that start with a 1, namely
    (1,0,0) and (1,1,1). There are 4 ways of starting with a 1 without
    regard to the number of 1's. There are 8 strings overall, so the
    probability p(odd)*p(start with 1) = (4/8) * (4/8) while the
    probability p(odd and start with 1) = 2/8. These are the same
    fractions, so the 2 events are independent.

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