Math 340                                                        
	   Remarks and Hints Related to Homework Problems
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1.  (page 91,  #3)

    This problem is equivalent to rolling one die 5 times.  Thus, use
    n = 5, p=P(S)=1/6,  and  X= no of 6's in 5 rolls.  Then,  for example

    P(C) = P(X < or = 2),  etc.
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2.  (p.91,  #5)

     In this problem  n = 20 and p = 1/2.  Note that we are GIVEN  that
     20 heads have occurred in 20 tosses.  So,  all probabilities should
     be treated as conditional ones,  namely,

	    P(A|B) = P(AB) / P(B).

     a.  Let  A = the event that the first toss landed heads
	      B = the event that there were 12 heads in 20 tosses
	      X = the number of heads in 20 tosses.
  
	 First of all note that the distribution of X is B(20, 1/2).
         Then,

	 P(B) = P(X=12) = (20 choose 12) (1/2)^12 (1/2)^8
	      = (20 nCr 12) (1/2)^20.

	 P(AB) = P(A) P(there were 11 heads in the remaining 19 tosses)
	       = (1/2) [(19 nCr 11) (1/2)^19]
	       = (19 nCr 11) (1/2)^20

	 Hence,

	 P(A|B) = [(19 nCr 11) (1/2)^20]/[(20 nCr 12) (1/2)^20]
		= (19 nCr 11) / (20 nCr 12) 

     b.  Let  C = the event that the first two tosses landed heads

     c.  Let  D = the event that AT LEAST two of the first five tosses 
		  landed heads.
        
	Thus, we want P(D|B).

	Note that D means that there COULD have been 2 or 3 or 4 or 5
	 heads in the first 5 tosses. [It may be easier to consider the 
	 complementary event in this case.] Below, we will actually 
	 calculate 1 - P(D^c|B).

	 (D^c AND B) = the event that the number of heads observed in the first
	               five tosses is 0 or 1, given tht there were 20 heads.
         So,

	 P(no heads in the first 5 tosses AND 12 heads in 20 tosses)
	   = P(no heads in the first tosses) *
	     P(12 heads in the remaining 15 tosses)
           = (1/2)^5 [(15 nCr 12) (1/2)^15]
	   = (15 nCr 12) (1/2)^20
         Also,

	 P(one head in the first 5 tosses AND 11 heads in 20 tosses)
	   = P(one head in the first 5 tosses) *
	     P(11 heads in the remaining 15 tosses)
           = [(5 nCr 1) (1/2)^5] [(15 nCr 11) (1/2)^15]
	   = 5 (15 nCr 11) (1/2)^20
         Therefore,

	 P(D^c AND B) = (15 nCr 12) (1/2)^20 + 5 (15 nCr 11) (1/2)^20

	 Hence,

	 P(D^c | B) = P(D^c AND B) / P(B)
		    = [(15 nCr 12)/(20 nCr 12)] + [5 (15 nCr 11)/(20 nCr 12)]

	 And finally the answer to the problem is 1-P(D^c|B).
		   
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3.  (p.91,  #7)

     Let,  A = the number that my die shows after a roll
	   B = the number that your die shows after a roll
	   X = the number of times that you win in this game

     Note that n=5 in this problem and we are looking for P(X >= 4).
     Thus,  the only thing that is missing is p=P(S)=P(A < B).

     For example you would win if (A=5 AND B=6). Hence,

     P(A < B) = sum_{i=1}^{5} sum_{j=2, j>i}^{6} P(A=i, B=j).

     The double sum above can easily be evaluated in three ways:

     a. we note that P(A=i, B=j) = 1/36 for all i,j=1,...,6

     b. we note that P(A=i, B=j) = P(B=j | A=i) P(A=i) = (1/6)(1/6)

     c. symmetry argument. See if you can figure this one out.
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4.  (p.91,  #11)

     n=15,  p=P(S)=0.7
     Define,  X = the number of adults in the sample.

     a.  What is the EXPECTED number of adults in the sample?
     b.  What is the probability that X is equal to your anser in (a)?
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