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Solution to Exercise 3.62.

(a)

$\begin{displaymath} P[X = 2] \; = \; \frac{ \left( \begin{array} {c} 6 \\ 2 \en... ...t) }{ \left( \begin{array} {c} 15 \\ 5 \end{array} \right) } .\end{displaymath}$

Now

$\begin{displaymath} \left( \begin{array} {c} 6 \\ 2 \end{array} \right) \; = \; \frac{6 * 5}{2} \; = \; 15 ,\end{displaymath}$

and

$\begin{displaymath} \left( \begin{array} {c} 9 \\ 3 \end{array} \right) \; = \; \frac{ 9 * 8 * 7 }{ 2 * 3 } \; = \; 84 ,\end{displaymath}$

and

$\begin{displaymath} \left( \begin{array} {c} 15 \\ 5 \end{array} \right) \; = \; \frac{ 15 * 14 * 13 * 12 * 11}{2 * 3 * 4 * 5} \; = \; 3003 .\end{displaymath}$

Finally,

$\begin{displaymath} P[X = 2] \; = \; \frac{15 * 84}{3003} \; = \; 0.4195804 .\end{displaymath}$

(b)

$\begin{displaymath} P[ X = 1 ] \; = \; \frac{ \left( \begin{array} {c} 6 \\ 1 ... ...t) }{ \left( \begin{array} {c} 15 \\ 5 \end{array} \right) } .\end{displaymath}$

Now

$\begin{displaymath} \left( \begin{array} {c} 6 \\ 1 \end{array} \right) \; = \; 6 ,\end{displaymath}$

$\begin{displaymath} \left( \begin{array} {c} 9 \\ 4 \end{array} \right) \; = \; \frac{9 * 8 * 7 * 6}{2 * 3 * 4} \; = \; 126 .\end{displaymath}$

$\begin{displaymath} P[X = 1] \; = \; \frac{6 * 126}{3003} \; = \; 0.2517483 .\end{displaymath}$

Also we need to compute

$\begin{displaymath} P[ X = 0 ] \; = \; \frac{ \left( \begin{array} {c} 6 \\ 0 ... ...t) }{ \left( \begin{array} {c} 15 \\ 5 \end{array} \right) } .\end{displaymath}$

We have

$\begin{displaymath} \left( \begin{array} {c} 6 \\ 0 \end{array} \right) \; = \; 1 ,\end{displaymath}$

$\begin{displaymath} \left( \begin{array} {c} 9 \\ 5 \end{array} \right) \; = \... ...4 * 5} \; = \; \frac{9 * 8 * 7 *6 }{2 * 3 * 4 } \; = \; 126 .\end{displaymath}$

Hence,

$\begin{displaymath} P[ X = 0 ] \; = \; 126/3003 \; = \; 0.04195804 .\end{displaymath}$

Putting it all together,

$\begin{displaymath} P[ X \le 2 ] \; = \; P[ X = 0 ] + P[ X = 1 ] + P[ X = 2 ] \; = \; 0.4195804 + 0.2517483 + 0.04195804 \; = \; 0.7132867 .\end{displaymath}$

(c)

$\begin{displaymath} P[ X \ge 2 ] \; = \; 1 - P[ X \le 1 ] \; = \; 1 - \left\{ P... ...\} \; = \; 1 - ( 0.2517483 + 0.04195804 ) \; = \; 0.7062937 .\end{displaymath}$

(d)

$\begin{displaymath} E[X] \; = \; 5 * \frac{6}{15} \; = \; 2 .\end{displaymath}$

$\begin{displaymath} V[X] \; = \; \left( \frac{15 - 5}{15 -1} \right) * 5 * \frac{6}{15} * \left( 1 - \frac{6}{15} \right) \; = \; 0.8571429 .\end{displaymath}$

Dennis Cox
2/19/2001