A video game has a large community of online players. Before starting a game, the player must choose between two ``worlds'': either world A or world B.
An individual is chosen at random from the community of players.
When playing a game, we assume that:
\begin{itemize}
  \item the probability that the player chooses world A is equal to $\frac{2}{5}$;
  \item if the player chooses world A, the probability that they win the game is $\frac{7}{10}$;
  \item the probability that the player wins the game is $\frac{12}{25}$.
\end{itemize}
We consider the following events:
\begin{itemize}
  \item A: ``The player chooses world A'';
  \item B: ``The player chooses world B'';
  \item G: ``The player wins the game''.
\end{itemize}

This exercise is a multiple choice questionnaire (5 questions). For each question, only one of the four proposed answers is correct.

\begin{enumerate}
  \item The probability that the player chooses world A and wins the game is equal to:\\
a. $\frac{7}{10}$\\
b. $\frac{3}{25}$\\
c. $\frac{7}{25}$\\
d. $\frac{24}{125}$
  \item The probability $P_{B}(G)$ of event $G$ given that $B$ is realized is equal to:\\
a. $\frac{1}{5}$\\
b. $\frac{1}{3}$\\
c. $\frac{7}{15}$\\
d. $\frac{5}{12}$
\end{enumerate}

In the rest of the exercise, a player plays 10 successive games. This situation is treated as a random draw with replacement. We recall that the probability of winning a game is $\frac{12}{25}$.

3. The probability, rounded to the nearest thousandth, that the player wins exactly 6 games is equal to:\\
a. 0.859\\
b. 0.671\\
c. 0.188\\
d. 0.187

4. We consider a natural number $n$ for which the probability, rounded to the nearest thousandth, that the player wins at most $n$ games is 0.207. Then:\\
a. $n = 2$\\
b. $n = 3$\\
c. $n = 4$\\
d. $n = 5$

5. The probability that the player wins at least one game is equal to:\\
a. $1 - \left(\frac{12}{25}\right)^{10}$\\
b. $\left(\frac{13}{25}\right)^{10}$\\
c. $\left(\frac{12}{25}\right)^{10}$\\
d. $1 - \left(\frac{13}{25}\right)^{10}$