The function $f$ is defined on the interval $[ 0 ; 1 ]$ by:
$$f ( x ) = \frac { 0,96 x } { 0,93 x + 0,03 }$$

The fight against doping involves carrying out anti-doping tests which aim to determine whether an athlete has used prohibited substances. During a competition bringing together 1000 athletes, a medical team tests all competitors. We propose to study the reliability of this test.

Let $x$ denote the real number between 0 and 1 which represents the proportion of doped athletes. During the development of this test, it was possible to determine that:
\begin{itemize}
  \item the probability that an athlete is declared positive given that they are doped is equal to 0.96;
  \item the probability that an athlete is declared positive given that they are not doped is equal to 0.03.
\end{itemize}

We denote:
\begin{itemize}
  \item D the event: ``the athlete is doped''.
  \item $T$ the event: ``the test is positive''.
\end{itemize}

\begin{enumerate}
  \item Copy and complete the probability tree.
  \item Determine, as a function of $x$, the probability that an athlete is doped and has a positive test.
  \item Prove that the probability of event $T$ is equal to $0,93 x + 0,03$.
  \item For this question only, assume that there are 50 doped athletes among the 1000 tested. Prove that the probability that an athlete is doped given that their test is positive is equal to $f ( 0,05 )$. Give an approximate value rounded to the nearest hundredth.
  \item The positive predictive value of a test is called the probability that the athlete is truly doped when the test result is positive.
  \begin{enumerate}
    \item[a.] Determine from which value of $x$ the positive predictive value of the test studied will be greater than or equal to 0.9. Round the result to the nearest hundredth.
    \item[b.] A competition official decides to no longer test all athletes, but to target the most successful athletes who are assumed to be more frequently doped. What is the consequence of this decision on the positive predictive value of the test? Argue using a result from Part A.
  \end{enumerate}
\end{enumerate}