5. At a fair, there is a very inviting game, as it can be played for free; the winner gets a prize. The agreed prize for the first four games is one coin, for the fifth it is two coins. Nicola has a probability of $\frac{2}{3}$ of winning the prize in each game and decides to play 5 games. What is the probability that Nicola wins at least 4 coins?
(A) $5\left(\frac{2}{3}\right)^{5}$
(B) $4\left(\frac{2}{3}\right)^{5}$
(C) $3\left(\frac{2}{3}\right)^{5}$
(D) $2\left(\frac{2}{3}\right)^{5}$
(E) $\left(\frac{2}{3}\right)^{5}$.
5. The answer is (A). Let's consider separately the cases where Nicola has respectively lost and won the last game. In the first case (which occurs with probability $1 / 3$), Nicola must have won the first four games, an event that has a probability of $\left(\frac{2}{3}\right)^{4}$. In the second case (which occurs with probability $2 / 3$), Nicola must not have lost more than two games out of the first four: this latter event has a complementary probability to that of losing all the first four games, or winning exactly one out of four. It follows that the sought probability is:

$$
\frac{1}{3} \cdot\left(\frac{2}{3}\right)^{4}+\frac{2}{3}\left(1-\left(\frac{1}{3}\right)^{4}-4 \cdot \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{3}\right)=5 \cdot\left(\frac{2}{3}\right)^{5}=\frac{160}{243} .
$$