$\begin{array}{1 1}(A)\;0.51\\(B)\;0.61\\(C)\;0.41\\(D)\;0.61\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:

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

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

$\Rightarrow 0.31+0.20$

$\Rightarrow 0.51$

Hence (A) is the correct answer.

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