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A team of medical students doing their internship have to assist during surgeries at a city hospital.The probabilities of surgeries rated as very complex,complex,routine,simple or very simple are respectively 0.15,0.20,0.31,0.26,0.08.Find the probabilities that a particular surgery will be rated neither very complex nor very simple

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

1 Answer

  • $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Step 1:
Let us denote the surgeries very complex -A,complex -B,routine-C,simple -D,very simple -E
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.
answered Jul 7, 2014 by sreemathi.v

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