Incertidumbre 01

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\newcommand{\mahteacher}{Dr. Leonardo Garrido}
\newcommand{\mahclass}{Sistemas de Incertidumbre}
\newcommand{\mahtitle}{Tarea 01}
\newcommand{\mahdate}{20 de enero de 2016}
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\lhead{1170065 - Xavier Sánchez}
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\title{
    \vspace{1in}
    \textbf{Tecnológico de Monterrey} \\
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    \textmd{\mahclass} \\
    \large{\textit{\mahteacher}} \\
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    \textsc{\mahtitle}
    \author{01170065  - MIT \\
    Xavier Fernando Cuauhtémoc Sánchez Díaz \\
    \texttt{xavier.sanchezdz@gmail.com}}
    \date{\mahdate}
}

\begin{document}

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\begin{homeworkProblem}
    \textit{Corresponde al ejercicio \textbf{13.7} del libro.}

    \spacepls

    Consider the set of all possible five-card poker hands dealt fairly from a standard deck
    of fifty-two cards.

    \begin{enumerate}[label= \textbf{\alph*)}]
        \item How many atomic events are there in the joint probability distribution (i.e., how many
    five-card hands are there)?
        \item What is the probability of each atomic event?
        \item What is the probability of being dealt a royal straight flush? Four of a kind?
    \end{enumerate}

    \spacepls
    \solution
    \spacepls

    \textbf{a)} El número de combinaciones de 5 cartas posibles en un mazo de 52 cartas está dado por:

    \[52 \times 51 \times 50 \times 49 \times 48 = 311875200\]

    Esto es considerando que recibir \textbf{A, K, Q, J, 10} es un evento distinto de recibir \textbf{10, J, Q, K, A} por el orden. Si lo que importa son las "manos" distintas, entonces el número de combinaciones hay que dividirlo entre 5!, que es el número de "maneras distintas" en las que podemos recibir una sola combinación:

    \[\dfrac{311875200}{5!} = \dfrac{311875200}{120} = 2598960\]

    \textbf{b)} Cada evento es equiprobable, por lo que la probabilidad de cada uno es \(\dfrac{1}{311875200}\) si el orden importase, y  \(\dfrac{1}{2598960}\)  si lo que importa es la mano como tal.

    \spacepls

    \textbf{\(c_1\))} La probabilidad de obtener una flor imperial \textbf{del mismo palo} está dada por:

    \[P(A \wedge K \wedge Q \wedge J \wedge 10) = \dfrac{1}{52} \times \dfrac{1}{51} \times \dfrac{1}{50} \times \dfrac{1}{49} \times \dfrac{1}{48} = \dfrac{1}{2598960}\]

    Como hay 4 palos en la baraja, la probabilidad de obtener una flor imperial de cualquier palo es de \(\dfrac{4}{2598960} = \dfrac{1}{649740}\).

    \spacepls

    \textbf{\(c_2)\)} La probabilidad de obtener un poker (\textit{four of a kind}) está dada por:

    \[\dfrac{13 \times 4 \times 12}{2598960} = \dfrac{624}{2598960} = \dfrac{1}{4165}\]

    donde el 13 es cualquier carta de la baraja, 4 el número de cartas del mismo tipo que se precisa tener, y 12 posibilidades de las cartas restantes, todo en la primera mano.

\end{homeworkProblem}

\pagebreak

\begin{homeworkProblem}
 \textit{Corresponde al ejercicio \textbf{13.8} del libro.}

 \spacepls

 Given the full joint distribution shown in Figure 13.3, calculate the following:

 \begin{enumerate}[label={\alph*)}]
     \item \textbf{P}(\textit{toothache})
     \item \textbf{P}(\textit{Cavity})
     \item \textbf{P}(\textit{Toothache \(\mid\) cavity})
     \item \textbf{P}(\textit{Cavity \(\mid\) toothache \(\vee\) catch})
 \end{enumerate}

\begin{table}[h]
    \centering
    \begin{tabular}{l|l|l|l|l|}
           & \multicolumn{2}{l|}{toothache} & \multicolumn{2}{l|}{\(\neg\) toothache} \\ \hline
           & catch & \(\neg\) catch & catch & \(\neg\) catch \\\hline
    cavity & 0.108 & 0.012 & 0.072 & 0.008 \\
    \(\neg\) cavity & 0.016 & 0.064 & 0.144 & 0.576 \\
    \end{tabular}
\end{table}

 \spacepls
 \solution
 \spacepls

 \textbf{a)} \( \textbf{P}(toothache) = 0.108 + 0.016 + 0.012 + 0.064 = 0.2 \)

 \spacepls

 \textbf{b)} \(\textbf{P}(Cavity) = \langle 0.108 + 0.012 + 0.072 + 0.008, 0.016 + 0.064 + 0.144 + 0.576 \rangle  = \langle 0.2 , 0.8 \rangle\)

 \spacepls

 \textbf{c)} \(\textbf{P}(Toothache \mid cavity) = \dfrac{P(toothache \wedge cavity)}{P(cavity)}\), y \(P(cavity) = 0.108 + 0.012 + 0.072 + 0.008 = 0.2)\). Por tanto:

 \[ \textbf{P}(Toothache \mid cavity) = \left \langle \dfrac{0.108, 0.012}{0.2} , \dfrac{0.072, 0.008}{0.2} \right \rangle  = \langle 0.6, 0.4 \rangle\]

 \spacepls

 \textbf{d)} \(\textbf{P}(Cavity \mid toothache \vee catch) = \left \langle \dfrac{P(cavity \wedge toothache \vee catch)}{P(toothache \vee catch)} , \dfrac{P(\neg cavity \wedge toothache \vee catch)}{P(toothache \vee catch)} \right \rangle\), y

 \spacepls

 \(P(toothache \vee catch) = 0.108 + 0.012 + 0.016 + 0.064 + 0.072 + 0.144 = 0.416\), entonces:

 \[ \textbf{P}(Cavity \mid toothache \vee catch) = \left \langle \dfrac{0.108 + 0.012 + 0.072}{0.416} , \dfrac{0.016 + 0.064 + 0.144}{0.416} \right \rangle \approx \langle 0.4615, 0.5384 \rangle \]


\end{homeworkProblem}

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