Question 1035919
Let X = # times tails appear.

If the die turns up 2,{{{ p(2) = (1/6)*(1/4) = 1/24}}}.

If the die turns up 3, {{{p(3) = (1/6)*(1 - 3C0/2^3 - 3C1/2^3) = (1/6)(1/2) = 1/12}}}.

If the die turns up 4, {{{p(4) = (1/6)*(1 - 4C0/2^4 - 4C1/2^4) = (1/6)(11/16) = 11/96}}}.

If the die turns up 5, {{{p(5) = (1/6)*(1 - 5C0/2^5 - 5C1/2^5) = (1/6)(13/16) = 13/96}}}.

If the die turns up 6, {{{p(5) = (1/6)*(1 - 6C0/2^6 - 6C1/2^6 - 6C6/2^6) = (1/6)(7/8) = 7/48}}}.

Since all these cases are mutually exclusive, 

==> {{{p(2 <= X <= 5) = 1/24 + 1/12 + 11/96 + 13/96 + 7/48 = highlight(25/48)}}}