Question 1140149
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There are  {{{C[100]^2}}} = {{{(100*99)/2}}} = 4950 different pairs to form of 100 tickets (without looking on ordering inside these pairs).


It is the space of all samples.



Of them, exactly  4*(100-4) = 4*96 = 384 different pairs contain precisely one of 4 prize tickets.



So, the probability under the question is  {{{384/4950}}} = 0.07758 = 7.758%.
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