Question 1123139
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There are  {{{C[80]^3}}} = {{{(80*79*78)/(1*2*3)}}} = 82160 different ways to select three persons of 80.


If none of the three would buy a car from the friend, it means that all three are from the subset of 80-28 = 52 persons.


There are  {{{C[52]^3}}} = {{{(52*51*50)/(1*2*3)}}} = 22100 different ways to select three persons of 52.


The probability under the question is the ratio of these two integers:


    P = {{{22100/82160}}} = 0.2690 = 26.90%  (approximately).
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Solved.