Math 340                                                        
	   Remarks and Hints Related to Homework Problems
___________________________________________________________________________

Below for the complement of an event A I will use A^c, and for the intersection
of two events A and B I will either use AB or (A int B).
____________________________________________________________________________

(p.10, #4)
    a. P(LL) = P(0 or 00) = 2/38.
    b. P(at least one of us wins) = P(LW) + P(WL) + P(WW)  
				  = 18/38 + 18/38 + 0  
				  = 36/38  or,
                                  
				  = 1 - P(LL)
				  = 1 - 2/38
				  = 36/38, same number.
    
    c. P(at least one of us loses) = 1 - P(WW) = 1 - 0 = 1, or

				   = P(LW) + P(WL) + P(LL)
				   = 18/38 + 18/38 + 2/38
				   = 1.
______________________________________________________________________________

Page 10, #5

   a. 52 x 51 ordered pairs since we have 52 choices for the first card and
      51 choices for the second card.
   b. 4/52 since there are 4 aces and 52 cards in a deck of cards.
   c. It dependes on what the first card drawn was. So we need to consider

      P(ace on 2nd) = P(ace on 1st and ace on2nd) + P(not ace on 1st and ace
                                                      on 2nd)
                    = (4/52)(3/51) + (48/52)(4/51)
   d. (4/52)(3/51)
   e. P(ace and ace) + P(ace and not ace) + P(not ace and ace)
      = (4/52)(3/51) + (4/52)(48/51) + (48/52)(4/51)
______________________________________________________________________________

(p.31, #8)
   a. P(AUB) = P(A) + P(B) - P(AB) = 0.8
   b. P(A^c) = 1 - P(A) = 0.4
   c. P(B^c) = 1 - P(B) = 0.6
   d. P(A^c B)
      Note that for any event B: 
      B=B int (Omega) = B int (A U A^c) = (B int A) U (B int A^c)
      Hence,

      P(B) = P(B int A) + P(B int A^c), since these two sets are disjoint

      Therefore,

      0.4 = 0.2 + P(B int A^c)  =>  P(B int A^c) = 0.2
   e. P(AUB^c) = P((A^c int B)^c) = 1 - P(A^c int B) = 1 - 0.2 = 0.8
   f. P(A^c B^c) P((AUB)^c) = 1 - P(AUB) = 1 - 0.8 = 0.2
______________________________________________________________________________

(p.31,  #9c)

    Consider the following (I will use F^c for the complement of F
    in Omega, etc.):
    
    F^c G^c H = (F U G)^c H
	      = (F U G U H^c)^c
    Thus, the prob. of the l.h.s. above is equal to the prob. of the 
    r.h.s. above.  Hence,

    P(F^c G^c H)  =  1 - P(F U G U H^c)
		  =  1 - [P(F)+P(G)+P(H^c)-P(FG)-P(FH^c)-P(GH^c)+P(FGH^c)]

    The only term above that is more challenging is the last one.  Prove the
    following:

	    (FGH^c) U (FGH) = FG.
    
    Then,  since the two terms on the l.h.s. are disjoint we get:

	    P(FGH^c) + P(FGH) = P(FG).

    Every term in the last equation is known except P(FGH^c) which is the 
    one we wanted.
______________________________________________________________________________

(Page 31, #13)

   Proof is by induction. First prove the formula for the case n=2. Then,
   assuming that the formula is true for n=k, prove the formula for n=k+1.
   In that case we have

   P(A_1 U A_2 U ... U A_(k+1)) \leq P(A_1 U ... U A_k) + P(A_(k+1))
                                \leq P(A_1) + ... + P(A_(k+1))

   using the result of the case for n=1 and the assumption for the case
   of n=k+1.

   ADDITIONAL DETAILS:

   STEP 1: SHOW TRUE WHEN N=2.

   P(A U B) = P(A) + P(B) - P(AB) < P(A) + P(B), since P(AB) \geq 0
   
   STEP 2: ASSUME TRUE WHEN N=K.

   Hence, by this assumption we assume that:

   P(A_1 U ... U A_K)  \leq  Sum_{i=1}^{i=k} P(A_i).

   STEP 3: INDUCTIVE STEP, SHOW TRUE FOR N = K+1.

   We need to show that P(A_1 U ... U A_(k+1))  \leq  Sum_{i=1}^{i=k+1} P(A_i)

   Denote the event (A_1 U ... U A_k) by B. Then, the left-hand-side of what we 

   need to show is:

   LHS = P(B U A_{k+1})

       \leq P(B) + P(k+1), by STEP 1

       = P(A_1 U ... U A_k) + P_{k+1}, backsubstituting for the event B

       \leq P(A_1) + ... + P(A_k) + P(A_{k+1}), by STEP 2. QED! 
_____________________________________________________________________________