This is just simple use of Binomial distribution.
Case $1$: If more than one person wears a tight tie, the probability is: $$P_1 = 1- P_{\text {No person wears a tight tie}} - P_{\text {One person wears a tight tie}}-P_{\text {Two wear a tight tie}} - P_{\text {Three wear a tight tie}} $$ $$=1-\sum_{i=0}^{3} (\binom {25}{i}(0.9)^{25-i}(0.1)^i) $$ $$=1- (\binom {25}{0}(0.9)^{25}(0.1)^{0})-(\binom {25}{1}(0.9)^{24}(0.1)^1)- (\binom {25}{2}(0.9)^{23}(0.1)^2)-(\binom {25}{3}(0.9)^{22}(0.1)^3)$$
We have considered wearing a tight tie as a success, the probability of which equals $0.1$.
Case $2$: If no one wears a tight tie, the probability is: $$P_2 = (0.9)^{25}$$
You are right in calculating the probability of at least one person wears a tight tie. Hope it helps.