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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 $\mathrm{var}(S)=\frac{(N-5)\mathrm{var}(R)}{(N-1)}.$
\bigskip
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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