Dissertação 6 - ITA - 2ª Fase Dia 1 - ITA 2025

Probabilidade:

• casos totais = $2^{\mathrm{n}-1}$
• casos pedidos = $\left(\begin{array}{c}\mathrm{n}-1\\ 3\end{array}\right)$

Logo, $\mathrm{p}\left(\mathrm{n}\right)=\frac{\left(\begin{array}{c}\mathrm{n}-1\\ 3\end{array}\right)}{2^{\mathrm{n}-1}}$.

Quer-se n tal que: $\mathrm{p}\left(\mathrm{n}-1\right)\le \mathrm{p}\left(\mathrm{n}\right)\ge \mathrm{p}\left(\mathrm{n}+1\right)$

(i)  $\mathrm{p}\left(\mathrm{n}-1\right)\le \mathrm{p}\left(\mathrm{n}\right)\Rightarrow \frac{\left(\begin{array}{c}\mathrm{n}-1\\ 3\end{array}\right)}{2^{\mathrm{n}-1}}\le \frac{\left(\begin{array}{c}\mathrm{n}\\ 3\end{array}\right)}{2^{\mathrm{n}}}\Rightarrow \left(\mathrm{n}-4\right)\le \frac{\left(\mathrm{n}-1\right)}{2}\Rightarrow 2\mathrm{n}-8\le \mathrm{n}-1\Rightarrow \mathrm{n}\le 7.$

(ii)  $\mathrm{p}\left(\mathrm{n}\right)\ge \mathrm{p}\left(\mathrm{n}+1\right)\Rightarrow \frac{\left(\begin{array}{c}\mathrm{n}-1\\ 3\end{array}\right)}{2^{\mathrm{n}-1}}\ge \frac{\left(\begin{array}{c}\mathrm{n}\\ 3\end{array}\right)}{2^{\mathrm{n}}}\Rightarrow \left(\mathrm{n}-3\right)\ge \frac{\mathrm{n}}{2}\Rightarrow 2\mathrm{n}-6\ge \mathrm{n}\Rightarrow \mathrm{n}\ge 6.$

Ou seja, $6\le \mathrm{n}\le 7$. Como $\mathrm{p}\left(6\right)=\frac{\left(\begin{array}{c}6-1\\ 3\end{array}\right)}{2^5}=\frac{10}{2^5}$ e  $\mathrm{p}\left(7\right)=\frac{\left(\begin{array}{c}7-1\\ 3\end{array}\right)}{2^6}=\frac{20}{2^6}$, temos $\mathrm{p}\left(6\right)=\mathrm{p}\left(7\right)$e, portanto, os possíveis valores são $\boxed{6 \mathrm{e} 7}$.