$\begin{array}{1 1}(A)\;0.68\\(B)\;0.78\\(C)\;0.48\\(D)\;0.18\end{array} $

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Step 1:

Given :

$P(e_1)=P(e_2)=0.08$

$P(e_3)=P(e_4)=P(e_5)=0.1$

$P(e_6)=P(e_7)=0.2$

$P(e_8)=P(e_9)=0.07$

$A=\{e_1,e_5,e_8\}$

$B=\{e_2,e_5,e_8,e_9\}$

Addition of all probabilities will be equal to 1

Step 2:

Probability of $\bar{B}$ from P(B)

$\Rightarrow P(\bar{B})=1-P(B)$

$\Rightarrow 1-0.32$

$\Rightarrow 0.68$

Step 3:

Probability of $\bar{B}$ from the elementary outcomes

$\Rightarrow$ Probability of ($e_1,e_2.........e_9)-P(e_2,e_5,e_8,e_9)$

$\therefore P(e_1,e_3,e_3,e_6,e_1)$

$\Rightarrow 0.08+0.1+0.1+0.2+0.2$

$\Rightarrow 0.68$

Hence (A) is the correct answer.

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