Question 1159182
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(A).  The probability of getting two heads and a six is  {{{(1/2)^2*(1/6)}}} = {{{1/24}}}.



(B)  The probability of getting one tail, one head and a three is  {{{2*(1/2)*(1/2)*(1/6)}}} = {{{1/12}}}.



(C)  The probability to get the event  (A or B) is the sum  {{{1/24}}} + {{{1/12}}} = {{{3/24}}} = {{{1/8}}}.
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