5. [4] A best of 9 series is to be played between two teams. That is, the first team to win 5 games is the winner. One of the teams, the Mathletes, has a $2 / 3$ chance of winning any given game. What is the probability that the winner is determined in the 7 th game?
Answer: $\frac{20}{81}$. If the Mathletes are the winners, they must win the 7 th game and have won exactly four of the previous 6 games. The probability of this occurring is
$$
\left((2 / 3)^{4} \cdot(1 / 3)^{2} \cdot\binom{6}{2}\right) \cdot(2 / 3) \text {. }
$$

Analogously, the other team wins with probability $\left((1 / 3)^{4} \cdot(2 / 3)^{2} \cdot\binom{6}{2}\right) \cdot(1 / 3)$. Summing, the probability is
$$
\frac{\binom{6}{2} \cdot 2^{2} \cdot\left(2^{3} \cdot 1^{2}+1^{5}\right)}{3^{7}}=\frac{5 \cdot 4}{3^{4}}=\frac{20}{81} .
$$