\textbf{Exercise 1 — Theme: Probability}\\
Results should be rounded if necessary to $10^{-4}$\\
A statistical study conducted in a company provides the following information:
\begin{itemize}
  \item $48\%$ of employees are women. Among them, $16.5\%$ hold a managerial position;
  \item $52\%$ of employees are men. Among them, $21.5\%$ hold a managerial position.
\end{itemize}
A person is chosen at random from among the employees. The following events are considered:
\begin{itemize}
  \item $F$: ``the chosen person is a woman'';
  \item $C$: ``the chosen person holds a managerial position''.
\end{itemize}
\begin{enumerate}
  \item Represent the situation with a probability tree.
  \item Calculate the probability that the chosen person is a woman who holds a managerial position.
  \item a. Prove that the probability that the chosen person holds a managerial position is equal to 0.191.\\
b. Are the events $F$ and $C$ independent? Justify.
  \item Calculate the probability of $F$ given $C$, denoted $P_{C}(F)$. Interpret the result in the context of the exercise.
  \item A random sample of 15 employees is chosen. The large number of employees in the company allows this choice to be treated as sampling with replacement.\\
Let $X$ be the random variable giving the number of managers in the sample of 15 employees.\\
Recall that the probability that a randomly chosen employee is a manager is equal to 0.191.\\
a. Justify that $X$ follows a binomial distribution and specify its parameters.\\
b. Calculate the probability that the sample contains at most 1 manager.\\
c. Determine the expected value of the random variable $X$.
  \item Let $n$ be a natural number. In this question, consider a sample of $n$ employees.\\
What must be the minimum value of $n$ so that the probability that there is at least one manager in the sample is greater than or equal to 0.99?
\end{enumerate}