$\begin{array}{1 1}(A)\;0.40\\(B)\;0.50\\(C)\;0.60\\(D)\;0.70\end{array} $

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- $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

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:

According to addition law probability

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$P(A)=\large\frac{n(e_1,e_5,e_8)}{n(S)}$

$\Rightarrow \large\frac{0.08+0.1+0.07}{1}$

$\Rightarrow 0.25$

Step 3:

$P(B)=\large\frac{n(e_2,e_5,e_8,e_9)}{n(S)}$

$\Rightarrow \large\frac{0.08+0.1+0.07+0.7}{1}$

$\Rightarrow 0.32$

Step 4:

$P(A \cap B)=P(e_1,e_5,e_8) \cap P(e_2,e_5,e_8,e_9)$

$\Rightarrow P(e_5,e_8)$

$\Rightarrow 0.1+0.07$

$\Rightarrow 0.17$

Step 5:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow 0.25+0.32-0.17$

$\Rightarrow 0.57-0.17$

$\Rightarrow 0.40$

Hence (A) is the correct answer.

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