(16 甲、乙两人进行乒乓球单打比赛, 采用五局三胜制 (即先胜三局者获冠军). 对于每局比赛, 甲获胜的概率为 $\frac{2}{3}$, 乙获胜的概率为 $\frac{1}{3}$. 如果将 “乙获得冠军” 的事件称为 “爆出冷门”. 试求此项赛事爆出冷门的概率.

(16 Two players, A and B, are playing a singles table tennis match, using a best-of-five format (i.e., the first to win three games wins the championship). For each game, the probability of A winning is $\frac{2}{3}$, and the probability of B winning is $\frac{1}{3}$. If the event of "B winning the championship" is referred to as "an upset". Try to find the probability of an upset in this match.
16 If one side wins with a score of $3: 2$, $3: 1$, or $3: 0$, the remaining game is added without affecting the match result. Thus, the problem is transformed into: finding the probability that player B wins at least three out of five games. The probability of B winning all five games is $\left(\frac{1}{3}\right)^{5}$; the probability of B winning four games and losing one is $\mathrm{C}_{5}^{1}\left(\frac{1}{3}\right)^{4} \times \frac{2}{3}$; the probability of B winning three games and losing two is $\mathrm{C}_{5}^{2}\left(\frac{1}{3}\right)^{3} \times\left(\frac{2}{3}\right)^{2}$. Adding these results, the probability that B wins at least three out of five games is $\frac{17}{81}$.