$\begin{array}{1 1}(A)\;0.35\\(B)\;0.45\\(C)\;0.55\\(D)\;0.65\end{array} $

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

Step 1:

Let us denote the surgeries very complex -A,complex -B,routine-C,simple -D,very simple -E

$P(A)=0.15$

$P(B)=0.20$

$P(C)=0.31$

$P(D)=0.26$

$P(E)=0.08$

Where by A,B,C,D & E are mutually exclusive to each other.

Step 2:

$P(B$ or $A)=P(A \cup B)$

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

$P(A \cap B)=\phi$

$\Rightarrow P(A)+P(B)$

$\Rightarrow 0.15+0.20$

$\Rightarrow 0.35$

Hence (A) is the correct answer.

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