$\begin{array}{1 1}(A)\;0.35\\(B)\;1.77\\(C)\;0.75\\(D)\;0.77\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:

$P(A'$ or $E')=P(A'\cup E')=P(A')+P(E')-P(A' \cap E')$

$P(A'\cap E')=\phi$

$\Rightarrow P(A')+P(E')$

$\Rightarrow (1-P(A))+(1-P(E))$

$\Rightarrow (1-0.15)+(1-0.08)$

$\Rightarrow 1.77$

Hence (B) is the correct answer.

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