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QUESTION

 In a sweepstake there are $N$ tickets ($N>5$), and five prizes are
   to be won. The prizes are allocated by drawing five tickets
   from a hat one after the other. Each winning ticket is replaced
   in the hat before the next ticket is drawn, so that a ticket
   may win more than one prize. The number of prizes won by a
   competitor who has brought two tickets is a random variable
   $R.$    State the value of $E(R)$ and show that
   $\mathrm{var}(R)=\frac{10(N-2)}{N^2}$.\\
   The rules are changed so that winning tickets are no longer
   replaced. The number of prizes won by a competitor with two
   tickets is $S.$

   Find $P(S=0),P(S=1),P(S=2).$

   Show that $E(S)=E(R)$ and $\mathrm{var}(S)=\frac{(N-5)\mathrm{var}(R)}{(N-1)}.$

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 ANSWER

 If he buys two tickets, probability of winning on
each draw is
   $\frac{2}{N}.\ \ R\sim B(5,\frac{2}{N})$ since the tickets are
   replaced.\\
   $\mu=5 \times \frac{2}{N}=\frac{10}{N}, \sigma ^2 =5\times
   \frac{2}{N}(1-\frac{2}{N})=\frac{10(N-2)}{N^2}$
   Without replacement the distribution is hypergeometric.

   \begin{eqnarray*}
    P(S=0)&=&\frac{\left(
    \begin{array}{c}N-2\\5\end{array}\right)\left(\begin{array}{c}2\\0\end{array}\right)}{\left(
    \begin{array}{c}N\\5\end{array}\right)}\\
    &=&\frac{(N-2)(N-3)(N-4)(N-5)(N-6)}{N(N-1)(N-2)(N-3)(N-4)}\\
    &=&\frac{(N-5)(N-6)}{N(N-1)}
   \end{eqnarray*}

   \begin{eqnarray*}
    P(S=1)&=&\frac{\left(
    \begin{array}{c}N-2\\4\end{array}\right)\left(\begin{array}{c}2\\1\end{array}\right)}{\left(
    \begin{array}{c}N\\5\end{array}\right)}\\
    &=&\frac{(N-2)(N-3)(N-4)(N-5)\times2 \times5}{4!N(N-1)(N-2)(N-3)(N-4)}\\
    &=&\frac{10(N-5)}{N(N-1)}
   \end{eqnarray*}

    \begin{eqnarray*}
    P(S=2)&=&\frac{\left(
    \begin{array}{c}N-2\\3\end{array}\right)\left(\begin{array}{c}2\\2\end{array}\right)}{\left(
    \begin{array}{c}N\\5\end{array}\right)}\\
    &=&\frac{(N-2)(N-3)(N-4)5!)}{3!N(N-1)(N-2)(N-3)(N-4)}\\
    &=&\frac{20}{N(N-1)}
   \end{eqnarray*}

   Check that these add to give 1.
    $$E(S)=\frac{10(N-5)}{N(N-1)}+\frac{40}{N(N-1)}=\frac{10}{N}$$
   \begin{eqnarray*}
   \textrm{Var}(S)&=&\frac{10(N-5)}{N(N-1)}+\frac{80}{N(n-1)}-\frac{100}{N^2}\\
   &=&\frac{10(N^2-7N+10)}{N^2(N-1)}\\&=&\frac{10(N-2)(N-5)}{N^2(N-1)}\\&=&
   \frac{(N-5)}{(N-1)}\textrm{Var}(R)
   \end{eqnarray*}


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