Question 1142661
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(a)  {{{C[24]^3}}} = {{{((24*23*22)/(1*2*3))}}} = 2024 different combinations of 3 cans.



(b)   {{{C[24-1]^2}}} = {{{C[23]^2}}} = {{{((23*22)/(1*2))}}} = 253 different combinations of 3 cans, containing the contaminated one.


      (You can form all possible combinations by adding any two other cans to the contaminated one).


      The probability is  {{{253/2024}}} = 0.125.
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