\section*{Part 1}
Julien must take the plane; he planned to take the bus to get to the airport.\\
If he takes the 8 o'clock bus, he is sure to be at the airport in time for his flight.\\
On the other hand, the next bus would not allow him to arrive at the airport in time.\\
Julien left late from his apartment and the probability that he misses his bus is 0.8.\\
If he misses his bus, he goes to the airport by taking a private car company; he then has a probability of 0.5 of being on time at the airport.\\
We denote:
\begin{itemize}
  \item $B$ the event: ``Julien manages to take his bus'';
  \item $V$ the event: ``Julien is on time at the airport for his flight''.
\end{itemize}

\begin{enumerate}
  \item Give the value of $P _ { B } ( V )$.
  \item Represent the situation with a probability tree.
  \item Show that $P ( V ) = 0.6$.
  \item If Julien is on time at the airport for his flight, what is the probability that he arrived at the airport by bus? Justify.
\end{enumerate}

\section*{Part 2}
Airlines sell more tickets than there are seats on planes because some passengers do not show up for boarding on the flight they have booked. This practice is called overbooking.\\
Based on statistics from previous flights, the airline estimates that each passenger has a 5\% chance of not showing up for boarding.\\
Consider a flight on a plane with 200 seats for which 206 tickets have been sold. We assume that the presence at boarding of each passenger is independent of other passengers and we call $X$ the random variable that counts the number of passengers showing up for boarding.

\begin{enumerate}
  \item Justify that $X$ follows a binomial distribution and specify its parameters.
  \item On average, how many passengers will show up for boarding?
  \item Calculate the probability that 201 passengers show up for boarding. The result should be rounded to $10 ^ { - 3 }$.
  \item Calculate $P ( X \leqslant 200 )$, the result should be rounded to $10 ^ { - 3 }$. Interpret this result in the context of the exercise.
  \item The airline sells each ticket for 250 euros.
\end{enumerate}

If more than 200 passengers show up for boarding, the airline must refund the plane ticket and pay a penalty of 600 euros to each affected passenger.\\
We call:\\
$Y$ the random variable equal to the number of passengers who cannot board despite having purchased a ticket;\\
$C$ the random variable that totals the revenue of the airline on this flight.

We admit that $Y$ follows the probability distribution given by the following table:

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$y _ { i }$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$P \left( Y = y _ { i } \right)$ & 0,94775 & 0,03063 & 0,01441 & 0,00539 & 0,00151 & 0,00028 &  \\
\hline
\end{tabular}
\end{center}

a. Complete the probability distribution given above by calculating $P ( Y = 6 )$.\\
b. Justify that: $C = 51500 - 850 Y$.\\
c. Give the probability distribution of the random variable $C$ in the form of a table. Calculate the expected value of the random variable $C$ to the nearest euro.\\
d. Compare the revenue obtained by selling exactly 200 tickets and the average revenue obtained by practicing overbooking.