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

A correct answer earns one point. An incorrect answer, a multiple answer, or the absence of an answer to a question earns or deducts no points.

\textbf{PART I}

In a mail processing centre, a machine is equipped with an automatic optical reader for recognizing postal addresses. This reading system correctly recognizes $97\%$ of addresses; the remaining mail, which will be described as unreadable for the machine, is directed to a centre employee responsible for reading the addresses.\\
This machine has just read nine addresses. We denote by $X$ the random variable that gives the number of unreadable addresses among these nine addresses.\\
We assume that $X$ follows the binomial distribution with parameters $n = 9$ and $p = 0.03$.

\begin{enumerate}
  \item The probability that none of the nine addresses is unreadable is equal, to the nearest hundredth, to:\\
a. 0\\
b. 1\\
c. 0.24\\
d. 0.76
  \item The probability that exactly two of the nine addresses are unreadable for the machine is:\\
a. $\binom{9}{2} \times 0.97^{2} \times 0.03^{7}$\\
b. $\binom{7}{2} \times 0.97^{2} \times 0.03^{7}$\\
c. $\binom{9}{2} \times 0.97^{7} \times 0.03^{2}$\\
d. $\binom{7}{2} \times 0.97^{7} \times 0.03^{2}$
  \item The probability that at least one of the nine addresses is unreadable for the machine is:\\
a. $P(X < 1)$\\
b. $P(X \leqslant 1)$\\
c. $P(X \geqslant 2)$\\
d. $1 - P(X = 0)$
\end{enumerate}

\textbf{PART II}

An urn contains 5 green balls and 3 white balls, indistinguishable to the touch.\\
We draw at random successively and without replacement two balls from the urn.\\
We consider the following events:
\begin{itemize}
  \item $V_{1}$: "the first ball drawn is green";
  \item $B_{1}$: "the first ball drawn is white";
  \item $V_{2}$: "the second ball drawn is green";
  \item $B_{2}$: "the second ball drawn is white".
\end{itemize}

\begin{enumerate}
  \setcounter{enumi}{3}
  \item The probability of $V_{2}$ given that $V_{1}$ is realized, denoted $P_{V_{1}}\left(V_{2}\right)$, is equal to:\\
a. $\frac{5}{8}$\\
b. $\frac{4}{7}$\\
c. $\frac{5}{14}$\\
d. $\frac{20}{56}$
  \item The probability of event $V_{2}$ is equal to:\\
a. $\frac{5}{8}$\\
b. $\frac{5}{7}$\\
c. $\frac{3}{28}$\\
d. $\frac{9}{7}$
\end{enumerate}