At the outset, we consider the sums a+b+c of all 3 element subsets a,b,c of S = {1,2,......,63}. Divide the class of all 3 element subsets of S into four classes I, II, III, IV according as their sums are less than 95, greater than 97, equal to 96 or 97 and equal to 95 respectively.

To each A = {a,b,c} of Class I, associate a subset :
A1 = {a1, b1, c1}, where a1 = 64 -a; a2 = 64 -b and a3 = 64 -c
Then, a1+b1+c1 = 192 - (a+b+c) gt 97.

Hence, A1 belongs to class II.

It is clear that distinct sets in class I are associated with distinct sets in class II. So, class II contains contains as many members as Class I. Class II is not empty since {31, 32, 33} belongs to it.

Hence, the classes I and II together contain more subsets than class I.

Accordingly, the three element subsets (a,b,c) of (1,2,......, 62, 63) with a+b+c lt 95 is less than the number of those with a+b+c gt 95.

Consequently, probability that the sum of the three integers will be greater than 95 is more than the probability that the said sum would be less than 95.

NOTE: The symbol > and the symbol < are respectively denoted by gt and lt.

*** Also refer to the solution posted by Charlie in this location.

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