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Let $\xi$ be the number of successes of type $A$, $\eta$ be the number of successes of type $B$, and $n$ be the number of experiments in the series.\\
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It's trivial to show that both $\xi$ and $\eta$ have binomial distributions with $p$=$\frac{1}{3}$, so $\mbox{M}\xi=\mbox{M}\eta=np=\frac{n}{3}$, and $\mbox{D}\xi=\mbox{D}\eta=np\cdot(1-p)=\frac{2n}{9}$.\\
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Let's calculate $r(\xi,\eta)=\frac{\mbox{M}\xi\eta-\mbox{M}\xi\cdot\mbox{M}\eta}{\sqrt{\mbox{D}\xi\cdot\mbox{D}\eta}}$:\\
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\begin{displaymath}
\mbox{M}\xi\eta=\sum_{k=0}^n\sum_{i=0}^nik\cdot{}P(k=\xi,i=\eta)=\sum_{k=0}^n\sum_{i=0}^nkP(k=\xi)\cdot{}iP\mid_{k=\xi}(i=\eta)=
\end{displaymath}\begin{displaymath}
=\sum_{k=0}^nkP(k=\xi)\sum_{i=0}^niP\mid_{k=\xi}(i=\eta)
\end{displaymath}

$P\mid_{k=\xi}(i=\eta)$ equals zero when $k+i>n$, otherwise it is the probability of getting $i$ successes (of type $B$) out of series of $n-k$ experiments; each of those experiments has two equiprobable outcomes, namely $B$ and $C$. Therefore, when $k+i\le n$, this probability matches the probability of a binomially distributed variate to possess the value of $i$:
\begin{displaymath}
\sum_{k=0}^nkP(k=\xi)\sum_{i=0}^niP\mid_{k=\xi}(i=\eta)=\sum_{k=0}^nkP_{B(n,\frac{1}{3})}(k)\sum_{i=0}^{n-k}iP_{B(n-k,\frac{1}{2})}(i)=
\end{displaymath}\begin{displaymath}
=\sum_{k=0}^nkP_{B(n,\frac{1}{3})}(k)\cdot\mbox{M}_{B(n-k,\frac{1}{2})}=\sum_{k=0}^n\Big(kP_{B(n,\frac{1}{3})}(k)\cdot\frac{n-k}{2}\Big)=
\end{displaymath}\begin{displaymath}
=\frac{1}{2}\cdot\Big(\sum_{k=0}^nnkP_{B(n,\frac{1}{3})}(k) - \sum_{k=0}^nk^2P_{B(n,\frac{1}{3})}(k)\Big)=
\end{displaymath}\begin{displaymath}
=\frac{1}{2}\cdot\Big(n\cdot\mbox{M}_{B(n,\frac{1}{3})} - \big(\mbox{D}_{B(n,\frac{1}{3})}+{\mbox{M}^2}_{B(n,\frac{1}{3})}\big)\Big)=\frac{1}{2}\cdot\Big(n\cdot\frac{n}{3} - \frac{2n}{9}-\frac{n^2}{9}\Big)=
\end{displaymath}\begin{displaymath}
=\frac{1}{2}\cdot\Big(\frac{2n^2}{9}-\frac{2n}{9}\Big)=\frac{n^2}{9}-\frac{n}{9};
\end{displaymath}\begin{displaymath}
\mbox{cov}(\xi,\eta)=\mbox{M}\xi\eta-\mbox{M}\xi\cdot\mbox{M}\eta=\Big(\frac{n^2}{9}-\frac{n}{9}\Big)-\frac{n^2}{9}=-\frac{n}{9};
\end{displaymath}\begin{displaymath}
r=\mbox{cov}(\xi,\eta)\Bigg/\frac{2n}{9}=-\frac{1}{2}
\end{displaymath}
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