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Some answers to Sample Problems for Test 1

4.
a) S = $\{ $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),
                 (3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4) $\}$
b) $4/16 = 0.25$
5.
$P(A\cup B) = P(A) + P(B) - P(A\cap B) = 0.25 + 0.5 -0.05 = 0.70$
$P(A^c) = 1 - P(A) = 0.75$
$P(A^c \cup B^c) = P((A\cap B)^c) = 1 - P(A\cap B) = 1 - 0.05 = 0.95$
6.
a) $\frac{4}{10} \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$
b) $P(LR\ or\ RL) = P(LR)+P(RL)$, where $P(LR) = \frac{6}{10} \frac{4}{9} = \frac{24}{90}$ and $P(RL) =
\frac{6}{10} \frac{4}{9}=\frac{24}{90}$
7.
a) $P(RR) = \frac{1}{10}\frac{4}{10}$
b) $P(LR\ or\ RL) = P(LR)+P(RL)$, where $P(LR) = \frac{6}{10} \frac{4}{10}$ and $P(RL) = \frac{6}{10} \frac{4}{10}$
8.
a) $ \left(\begin{array}{c} 3 \\ 3 \end{array} \right) 0.45^{3} 0.55^0 = 0.45^{3}$
b) It is easier to state the problem in terms of the complement:
prob = $1 - \left( \left(\begin{array}{c} 10 \\ 0 \end{array} \right) 0.45^{0} 0.55^{10} + \left(\begin{array}{c} 10 \\ 1 \end{array} \right) 0.45^{1} 0.55^9 \right)$
9.
a) $26^{4}$
b) $\frac{26!}{(26-4)!} = 26 \cdot 25 \cdot 24 \cdot 23 $
c) $26 \cdot 25 \cdot 24 \cdot 23 \cdot 10 \cdot 9 \cdot 8$
10.
a) $\left(\begin{array}{c} 10 \\ 4 \end{array} \right)$
b) $\frac{30!}{(30-3)!} = 30 \cdot 29 \cdot 28$
11.
$\left(\begin{array}{c} 20 \\ 2 \end{array} \right) (\frac{1}{6} )^{2}(\frac{4}{6})^{18}$
12.
a) Let $X$= number of defective items in 50 produced. Set n=50, p=0.02, q=0.98.
Prob. at most one is defective = Prob. either none or one is defective =
$P(X \leq 1) = f(0) + f(1) = \left(\begin{array}{c} 50 \\ 0 \end{array} \right) ...
...0} + \left(\begin{array}{c} 50 \\ 1 \end{array} \right)
(0.02)^{1} (0.98)^{49}$.
b) Set $\mu = n p = 50 * 0.02 = 1$. As before, let $X$= number of defective items in 50 produced. Now,
Prob. At most one is defective = prob. either none or one is defective =
$f(0) + f(1)$ = $1^{0} e^{-1}/0! + 1^{1}/1! e^{-1} = 2 e^{-1}$.
13.
a) $P(X \leq 3 ) = P( \frac{X-4.0}{0.75} \leq \frac{3-4.0}{0.75} ) =
P(Z \leq -\frac{4}{3}) = \Phi(\frac{-4}{3}) = 1 - \Phi(\frac{4}{3})$ (see table)
b) $P(3 \leq X \leq 5) = P( \frac{3-4.0}{0.75} \leq \frac{X-4.0}{0.75} \leq \frac{5...
...\frac{4}{3} \leq Z \leq \frac{4}{3} ) = \Phi(\frac{4}{3}) - \Phi(\frac{-4}{3}) $.
c) $P(5 \leq X ) = P( \frac{5-4.0}{0.75} \leq \frac{X-4.0}{0.75} ) = P( \frac{4}{3}\leq Z) = 1-\Phi(\frac{4}{3})$.
14.
$P(58 \leq X \leq 62) = P( \frac{58-60}{1.5} \leq \frac{X-60}{1.5} \leq \frac{62...
...\frac{4}{3} \leq Z \leq \frac{4}{3}) = \Phi(\frac{4}{3}) -
\Phi(\frac{-4}{3}) $.





Orlando Merino
2000-10-02