A sample space consists of 9 elementary outcomes e_1, e_2, ...., …

MCQ

$A$ sample space consists of $9$ elementary outcomes $e_1, e_2, ...., e_9$ whose probabilities are
$P(e_1) = P(e_2 ) = 0.08, P(e_3 ) = P(e_4) = P(e_5) = 0.1$
$P(e_6) = P(e_7) = 0.2, P(e_8) = P(e_9) = 0.07$
Suppose $A = {e_1, e_5, e_8}, B = {e_2, e_5, e_8, e_9}$

A
Calculate $P (A), P (B),$ and $\text{P}(\text{A}\cap\text{B})$
B
Using the addition law of probability$,$ calculate $\text{P}(\text{A}\cup\text{B})$
C
List the composition of the event $\text{A}\cup\text{B},$ and calculate $\text{P}(\text{A}\cup\text{B})$ by adding the probabilities of the elementary outcomes.
D
Calculate $\text{P}(\bar{\text{B}})$ from $P(B),$ also calculate $\text{P}(\bar{\text{B}})$ directly from the elementary outcomes of $\bar{\text{B}}$

✓

Answer

$\text{P(A)}=\text{P}(\text{E}_1)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$
$\text{P(A)}=0.08+0.1+0.07=0.25$
$\text{P(B)}=\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$$\text{P(B)}=0.08+0.1+0.07+0.07=0.32$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
Now$, \text{A}\cap\text{B}=\big\{\text{E}_5,\text{E}_8\big\}$
$\therefore\ \text{P}(\text{A}\cap\text{B})=\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$
$=0.1+0.07=0.17$
$\therefore\ \text{P}(\text{A}\cup\text{B})=0.25+0.32-0.17=0.40$
$\text{A}\cup\text{B}=\big\{\text{E}_1,\text{E}_2,\text{E}_5,\text{E}_8,\text{E}_9\big\}$$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$
$=0.08+0.08+0.1+0.07+0.07=0.40$
$\because\ \text{P}\bar{(\text{B})}=1-\text{P(B)}$$=1-0.32=0.68$
and $\bar{\text{B}}=\big\{\text{E}_1,\text{E}_3,\text{E}_4,\text{E}_6,\text{E}_7\big\}$
$\therefore\ \text{P}(\bar{\text{B}})=\text{P}(\text{E}_1)+\text{P}(\text{E}_3)+\text{P}(\text{E}_4)+\text{P}(\text{E}_6)+\text{P}(\text{E}_7)$
$=0.08+0.1+0.1+0.2+0.2=0.68$