
Notons $G$ l'événement : \og L'élève gagne \fg.\\
Il y a en tout $44+33+40=117$ élèves sur l'ensemble des trois classes. On a alors l'arbre suivant :

\begin{center}
\begin{tikzpicture}[yscale=.8]
\node (omega) at (0,0) {$\bullet$};
\draw (omega) -- (2,2) node[right] (C1) {$C_1$} node[midway,above,scale=0.7] {$\frac{44}{117}$};
\draw (omega) -- (2,0) node[right] (C2) {$C_2$} node[midway,above,scale=0.7] {$\frac{33}{117}$};
\draw (omega) -- (2,-2) node[right] (C3) {$C_3$} node[midway,above,scale=0.7] {$\frac{40}{117}$};

\draw (C1) -- (4,2.5) node[right] (G1) {$G$} node[midway,above] {$\frac{1}{2}$};
\draw (C1) -- (4,1.5) node[right] (G1B) {$\overline{G}$};

\draw (C2) -- (4,0.5) node[right] (G2) {$G$} node[midway,above] {$\frac{1}{3}$};
\draw (C2) -- (4,-0.5) node[right] (G2B) {$\overline{G}$};

\draw (C3) -- (4,-1.5) node[right] (G3) {$G$} node[midway,above] {$\frac{1}{4}$};
\draw (C3) -- (4,-2.5) node[right] (G3B) {$\overline{G}$};
\end{tikzpicture}
\end{center}

On en déduit :
\[ p(G)=\frac{44}{117}\times\frac{1}{2}+\frac{33}{117}\times\frac{1}{3}+\frac{40}{117}\times\frac{1}{4}=\frac{43}{117}.\]
On sait de plus que :
\[ p_G\left(C_2\right)=\frac{p(C_2\cap G)}{p(G)}=\frac{\frac{33}{117}\times\frac{1}{3}}{\frac{43}{117}}=\frac{11}{43}.\]