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\begin{itemize}
 \item HW-5-1
  \begin{itemize}
   \item [(1)] P(an ace (1+1) occurs at least once in 5 rolls of the two dice)
               \begin{eqnarray*}
               &=& 1-P(\mbox{NO (1+1) occurs in every 5 rolls of the two dice}) \\
               &=& 1-\left(\frac{35}{36}\right)^5 \\
               &=& 0.13
               \end{eqnarray*}
   \item [(2)] Let X be the total sum occurs in a single roll of the pair of dice.\\
               The probability of a total sum of at least 4, but less than 11occurs in a
               single roll of the pair of dice is
               \begin{eqnarray*}
               \lefteqn{P(4\leq X < 11)}\\
               &=& P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+\\
               & & P(X=9)+P(X=10)\\
               &=& \frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}
                   +\frac{3}{36}+\frac{2}{36}\\
               &=& \frac{32}{36}\\
               &=& \frac{8}{9}
               \end{eqnarray*}
   \item [(3)]
    \begin{tabular}{c|ccccccccccc}
    $X$&2&3&4&5&6&7&8&9&10&11&12\\
    $P(X)$ & $\displaystyle\frac{1}{36}$ & $\displaystyle\frac{2}{36}$ &
    $\displaystyle\frac{3}{36}$ & $\displaystyle\frac{4}{36}$ & $\displaystyle\frac{5}{36}$ &
    $\displaystyle\frac{6}{36}$ & $\displaystyle\frac{5}{36}$ & $\displaystyle\frac{4}{36}$ &
    $\displaystyle\frac{3}{36}$ & $\displaystyle\frac{2}{36}$ & $\displaystyle\frac{1}{36}$\\
    \end{tabular}
    \end{itemize}

 \item HW-5-2
    \\
    \\
    \begin{tabular}{|l|l|l|l|}
    \hline
         & 40 or under & Over 40 & Total\\
    \hline
    Wearing a seat belt & 0.484 & 0.369 & 0.853 \\
    Not wearing seat belt & 0.066 & 0.081 & 0.147 \\
    \hline
    Total & 0.55 & 0.45 & 1\\
    \hline
    \end{tabular}
   \begin{itemize}
   \item [(2)]0.147
   \item [(3)]Let A=\{a driver stopped at random is not wearing a seat
   belt\}and B=\{the driver is over 40\}. The probability of the driver is over
   40 given a driver stopped at random is not wearing a seat belt
   is
   \begin{eqnarray*}
   P(B|A)&=&\frac{P(B \& A)}{P(A)} \\
         &=&\frac{0.081}{0.147} \\
         &=&0.55
   \end{eqnarray*}
   \item [(4)]0.55
  \end{itemize}

 \item HW-5-3
    \\
    \\
    Let X be the number of aces in 13 times. \\
    Then $X \sim Binominal (13, \displaystyle\frac{1}{6})$.\\
    \begin{eqnarray*}
    P(X=x)={13 \choose x}(\frac{1}{6})^x(1-\displaystyle\frac{1}{6})^{13-x}, x=0,1,\cdots,13.
    \end{eqnarray*}
    The probability if getting at least 2 but fewer than 7 aces is
    \begin{eqnarray*}
    P(2 \leq X < 7) &=& P(X=2)+P(X=3)+P(X+4)+P(X=5)+P(X=6)\\
                    &=& {13 \choose 2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{11}
                        + {13 \choose 3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{10}
                        + {13 \choose 4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{9}+\\
                    & & {13 \choose 5}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^{8}
                        + {13 \choose 6}\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^{7}\\
                    &=& 0.29+0.21+0.11+0.04+0.01\\
                    &=& 0.66.
    \end{eqnarray*}


\end{itemize}

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