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A sample space consists of 9 elementary outcomes $e_1, e_2$, $\ldots e_9$ whose probabilities are
$$
\mathrm{P}\left(\mathrm{e}_1\right)=\mathrm{P}\left(\mathrm{e}_2\right)=0.08, \mathrm{P}\left(\mathrm{e}_3\right)=\mathrm{P}\left(\mathrm{e}_4\right)=\mathrm{P}\left(\mathrm{e}_5\right)=0.1
$$
$$
P\left(e_6\right)=P\left(e_7\right)=0.2, P\left(e_8\right)=P\left(e_9\right)=0.07
$$
$\operatorname{Suppose} \mathrm{A}=\left\{\mathrm{e}_1, \mathrm{e}_5, \mathrm{e}_8\right\}, \mathrm{B}=\left\{\mathrm{e}_2, \mathrm{e}_5, \mathrm{e}_8, \mathrm{e}_9\right\}$,
(a) Calculate $P(A), P(B)$ and $P(A \cap B)$.
(b) Using the addition law of probability, calculate $\mathrm{P}(\mathrm{A}$ $\cap \mathrm{B})$
(c) List the composition of the event $A \cup B$, and calculate $\mathrm{P}(\mathrm{A} \cup \mathrm{B}$,$) by adding the probabilities of the$ elementary outcomes.
(d) Calculate $\mathrm{P}(\overline{\mathrm{B}})$ from $\mathrm{P}$ (B), also calculate $\mathrm{P}$
$(\bar{B})$ directly from the elementary outcomes of $\bar{B}$.
$$
\mathrm{P}\left(\mathrm{e}_1\right)=\mathrm{P}\left(\mathrm{e}_2\right)=0.08, \mathrm{P}\left(\mathrm{e}_3\right)=\mathrm{P}\left(\mathrm{e}_4\right)=\mathrm{P}\left(\mathrm{e}_5\right)=0.1
$$
$$
P\left(e_6\right)=P\left(e_7\right)=0.2, P\left(e_8\right)=P\left(e_9\right)=0.07
$$
$\operatorname{Suppose} \mathrm{A}=\left\{\mathrm{e}_1, \mathrm{e}_5, \mathrm{e}_8\right\}, \mathrm{B}=\left\{\mathrm{e}_2, \mathrm{e}_5, \mathrm{e}_8, \mathrm{e}_9\right\}$,
(a) Calculate $P(A), P(B)$ and $P(A \cap B)$.
(b) Using the addition law of probability, calculate $\mathrm{P}(\mathrm{A}$ $\cap \mathrm{B})$
(c) List the composition of the event $A \cup B$, and calculate $\mathrm{P}(\mathrm{A} \cup \mathrm{B}$,$) by adding the probabilities of the$ elementary outcomes.
(d) Calculate $\mathrm{P}(\overline{\mathrm{B}})$ from $\mathrm{P}$ (B), also calculate $\mathrm{P}$
$(\bar{B})$ directly from the elementary outcomes of $\bar{B}$.
Solution:
1174 Upvotes
Verified Answer
(a) Here, $P(A)=P\left(e_1\right)+P\left(e_5\right)+P\left(e_8\right)$
$=0.08+0.10+0.07=0.25$
$\& P(B)=P\left(e_2\right)+P\left(e_5\right)+P\left(e_8\right)+P\left(e_9\right)$
$=0.08+0.10+0.07+0.07=0.32$
Since $A \cap B=\left\{e_5, e_8\right\}$
$P(A \cap B)=P\left(e_5\right)+P\left(e_8\right)=0.10+0.07=0.17$
(b) $\quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.25+0.32-0.17=0.40$
(c) $(\mathrm{A} \cup \mathrm{B})=\left\{\mathrm{e}_1, \mathrm{e}_2, \mathrm{e}_5, \mathrm{e}_8, \mathrm{e}_9\right\}$
$\therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.08+0.08+0.10+0.07+0.07=0.40$
(d) $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-0.32=0.68$
$$
\begin{gathered}
\&(\overline{\mathrm{B}})=\left\{\mathrm{e}_1, \mathrm{e}_3, \mathrm{e}_4, \mathrm{e}_6, \mathrm{e}_7\right\} \\
\Rightarrow \quad \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}\left(\mathrm{e}_1\right)+\mathrm{P}\left(\mathrm{e}_3\right)+\mathrm{P}\left(\mathrm{e}_4\right)+\mathrm{P}\left(\mathrm{e}_6\right)+\mathrm{P}\left(\mathrm{e}_7\right) \\
=0.08+0.10+0.10+0.20+0.20=0.68
\end{gathered}
$$
$=0.08+0.10+0.07=0.25$
$\& P(B)=P\left(e_2\right)+P\left(e_5\right)+P\left(e_8\right)+P\left(e_9\right)$
$=0.08+0.10+0.07+0.07=0.32$
Since $A \cap B=\left\{e_5, e_8\right\}$
$P(A \cap B)=P\left(e_5\right)+P\left(e_8\right)=0.10+0.07=0.17$
(b) $\quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.25+0.32-0.17=0.40$
(c) $(\mathrm{A} \cup \mathrm{B})=\left\{\mathrm{e}_1, \mathrm{e}_2, \mathrm{e}_5, \mathrm{e}_8, \mathrm{e}_9\right\}$
$\therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.08+0.08+0.10+0.07+0.07=0.40$
(d) $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-0.32=0.68$
$$
\begin{gathered}
\&(\overline{\mathrm{B}})=\left\{\mathrm{e}_1, \mathrm{e}_3, \mathrm{e}_4, \mathrm{e}_6, \mathrm{e}_7\right\} \\
\Rightarrow \quad \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}\left(\mathrm{e}_1\right)+\mathrm{P}\left(\mathrm{e}_3\right)+\mathrm{P}\left(\mathrm{e}_4\right)+\mathrm{P}\left(\mathrm{e}_6\right)+\mathrm{P}\left(\mathrm{e}_7\right) \\
=0.08+0.10+0.10+0.20+0.20=0.68
\end{gathered}
$$
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