$\begin{array}{1 1}(A)\;0.67\\(B)\;0.57\\(C)\;0.87\\(D)\;\text{None of these}\end{array} $

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

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:

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

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

$P(C \cap D)=\phi$

$\Rightarrow 0.31+0.26$

$\Rightarrow 0.57$

Hence (B) is the correct answer.

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