Supplementary Exercises 4.122 and 4.123 of IPS7e
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Ratings of two wine testers on an ordinal scale from 1 to 5. The
combined distribution of ratings for the two tasters is given in tabular
form.


4.122:
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(a) The sample space is all 25 pairs of ratings for the two tasters:
    S=(1,1),(1,2),(1,3),...,(5,5). 
The probabilities are all >=0, and their sum equals 1. Therefore the
table of individual probabilities gives a valid probability distribution.


(b) The probability that the two tasters agree is obtained by summing the
probabilities for the values in the sample space that correspond to agreement
(actually, the diagonal values in the table):
    p(1,1)+p(2,2)+p(3,3)+p(4,4)+p(5,5) = 0.03+0.08+0.25+0.20+0.06 = 0.62.


(c) The probability that Taster 1 rates a wine higher than 3 is also obtained by
summing the (10) probabilities in question:
    p(4,1)+p(4,2)+...p(4,5)+p(5,1)+p(5,2)+...+p(5,5) = 0.00+0.02+...+0.02+0.06
                                                     = 0.39.
Similarly for Taster 2:
    p(1,4)+p(2,4)+...p(5,4)+p(1,5)+p(2,5)+...+p(5,5) = 0.00+0.02+...+0.02+0.06
                                                     = 0.39.

Therefore the two probabilities are equal. This occurs because the table 
of probabilities is in fact symmetrical (around the diagonal). It would have 
been perfectly valid for the table not to be symmetrical, as long as the
probability laws were met.


4.123:
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When the rating for Taster 1 was known to be 3, we are seeking a conditional 
probability for Taster 2 (conditional on the rating for Taster 1). 
Let T1 and T2 denote the ratings for the two tasters,
respectively. Then using the definition of conditional probability,

   P(T2>3|T1=3) = P(T2>3 and T1=3) / P(T1=3) 
                = (p(3,4)+p(3,5)) / (p(3,1)+p(3,2)+p(3,3)+p(3,4)+p(3,5))
                = (0.05+0.01) / (0.01+0.05+0.25+0.05+0.01) 
                = 0.06/0.37 = 6/37 = 0.162 ~= 16%.