A company calls people by telephone to sell them a product.

\begin{itemize}
  \item The company calls each person a first time:
  \item the probability that the person does not answer is equal to 0.6;
  \item if the person answers, the probability that they buy the product is equal to 0.3.
  \item If the person did not answer on the first call, a second call is made:
  \item the probability that the person does not answer is equal to 0.3;
  \item if the person answers, the probability that they buy the product is equal to 0.2.
  \item If a person does not answer on the second call, we stop contacting them.
\end{itemize}

We choose a person at random and consider the following events:\\
$D _ { 1 }$: ``the person answers on the first call'';\\
$D _ { 2 }$: ``the person answers on the second call'';\\
$A$: ``the person buys the product''.

\section*{Part A}
\begin{enumerate}
  \item Copy and complete the weighted tree opposite.
  \item Using the weighted tree, show that the probability of event $A$ is $P ( A ) = 0.204$.
  \item We know that the person bought the product. What is the probability that they answered on the first call?
\end{enumerate}

\section*{Part B}
We recall that, for a given person, the probability that they buy the product is equal to 0.204.

\begin{enumerate}
  \item We consider a random sample of 30 people. Let $X$ be the random variable that gives the number of people in the sample who buy the product.\\
a. We admit that $X$ follows a binomial distribution. Give, without justification, its parameters.\\
b. Determine the probability that exactly 6 people in the sample buy the product. Round the result to the nearest thousandth.\\
c. Calculate the expected value of the random variable $X$. Interpret the result.
  \item Let $n$ be a non-zero natural number. We now consider a sample of $n$ people. Determine the smallest value of $n$ such that the probability that at least one person in the sample buys the product is greater than or equal to 0.99.
\end{enumerate}