The probability is 0.6 that an “unfair” coin will turn up tails on any

The problem with your solution is that 1 or 2 tails can occur in several ways:

THH can occur in 3 ways: THH, HTH and HHT (basically it's # of arrangements of 3 letters THH out of whihc 2 H's are identical, which is 3!/2!=3). So, you should have P(THH)=0.6*0.4^2*3!/2!;
Similarly TTH can also occur in 3 way: TTH, THT and HTT. So, you should have P(TTH)=0.6^2*0.4*3!/2!.
Notice that we don't have the same issue with TTT, since it can occur only in one way: TTT.

\(P(T\geq{1})=0.6*0.4^2*\frac{3!}{2!}+0.6^2*0.4*\frac{3!}{2!}+0.6^3=0.936\).

Answer: E.

Though this question can be solved with an easier approach: \(P(T\geq{1})=1-P(H=3)=1-0.4^3=0.936\) (the probability of at least one tails is 1 minus the probability of zero tails).

Answer: E.

Hope it's clear.