Math 340                                                        
	   Remarks and Hints Related to Homework Problems
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1. (p.54, #5)

    This problem is pretty much like the one we did in class.
    So,  you want to start by defining certain events of
    significance.

    D = the event that a randomly chosen individual has this disease.  
    A = the event that the result of the test is positive,  i.e.,  the
	person is diagnosed as diseased.

    Then, we are given

    P(D) = 0.01
    P(A | D^c) = 0.05
    P(A^c | D) = 0.2

    a. We are asked to find P(A). Now, the result of a test can be positive
      in only two ways. One way is that we have the disease AND the test is
      positive, or we don't have the disease AND the test is positive.
      So, using the multiplication rule these JOINT probabilities can be 
      written as
    
      P(A) = P(A | D) P(D) + P(A | D^c) P(D^c)

      As you can see the only unknown in the above equation is P(A|D). Now,
      if we have the disease, then our test result can be either positive or
      negative. This (conditional) probability will be negative with prob 0.2,
      i.e., P(A^c|D)=0.2. Hence, it must be true that P(A|D) = 1 - 0.2 = 0.8.
      We can now compute P(A). 

    b.  P(A^c and D) = P(A^c | D) P(D) = (0.2)(0.01) = 0.002

    c.  P(A^c and D^c) = P(A^c | D^c) P(D^c) = (1 - 0.05)(0.99) = 0.9405

    d.  P(D | A) = P(A|D) P(D) / P(A), using Baye's rule
		 = (0.8)(0.01) / 0.0575
		 = 0.1391 or 14%.

    e.  Yes, these probabilities have empirical interpretations which means
	that they are accurate in a long-run setting.
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2. (p.54, #7)

   We are given 

     Prior probabilities  P(B_i) = 1/3, i=1, 2, 3
     Posterior prob       P(B_i | W) = (12/23) (i/(i+1)), i=1, 2, 3,
			  P(W) = 23/36.

   Now the key is to understand the strategy: "Pick the box with the
   highest posterior probability, given the observed color." So, we
   have to ask  ourselves which box to I pick if the observed color
   is white, and which box do I pick if the observed color is black?

   We already know the answer to the first part of the question. Namely,
   if the observed color is white, we pick B_3 since P(B_3 | W) is the
   highest prob amongst P(B_i |W). 

   But, we don't know the answer to the second part of our question.
   Namely, if the observed color is black, which box do we pick. So,
   we first need to find these posterior probabilities.

    The following is what we should consider in this problem:
    Strategy:  Choose the box with highest posterior prob.  given the
	       observed color.
    
    Let 

    B = the event that the observed color is black.

    Then

    P(B) = P(B | B_1) P(B_1) + ... + P(B | B_3) P(B_3)
	 = (1/2)(1/3) + (1/3)(1/3) + (1/4)(1/3) = 13/36

    Now we can compute the posterior prob P(B_i | B).

    P(B_i | B) = P(B|B_i)P(B_i) / P(B) implies

    P(B_1|B) = (1/2)(1/3)/(13/36) = 6/36, similarly
    P(B_2|B) = 4/13
    P(B_3|B) = 3/13

    Thus, we now know that if the observed color is black we should pick B_1.
    Finally, we can answer the question. Note that we are "correct" if either
    one of the following two disjoint events occur:

    "we are correct" = (B and B_1) or (W and B_3).

    Hence,

    P(we are correct) = P(B B_1) + P(W B_3)
		      = P(B_1 | B) P(B) + P(B_3 | W) P(W)
		      = (6/13)(13/36) + (9/23)(23/36)
		      = (1/6) + (1/4)
		      = 5/12.

 b. No, because we have optimized already our probability for success.

 c. Repeat (a) with the new priors. You will have to do it for both W and B.

 d. You will consistently pick box 1 because under the revised priors I would
    pick B_1 twice as many times as B_2 or B_3 (on the average).
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