A team of medical students doing their internship have to assist during surgeries as a city hospital. The probabilities of surgeries rated as very complex, complex, routine simple or very simple respectively, 0.15, 0.20, 0.31, 0.26 and 0.08. Find the probabilities that a particular surgery will be rated.

(i) Complex or very complex.

(ii) Neither very complex nor very simple.

(iii) Routine or complex.

(iv) Routine or simple.

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Solution

Let Let $E_{1}, E_{2}, E_{3}, E_{4}$ and $E_{5}$ be the event that surgeries are rated as very complex, complex, routine, simple or very simple, respectively then according to the given information:
$P (E_{1}) = 0.15, P (E_{2}) = 0.2, P (E_{3}) = 0.31, P (E_{4}) = 0.26, P (E_{5}) = 0.08$
(i) To find the probabilities that a particular surgery will be rated complex or very complex:
$\Rightarrow P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2})$
$= 0.15 + 0.2 - 0$
$∴ P (E_{1} \cup E_{2}) = 0.35$
(ii) To find the probabilities that a particular surgery will be rated neither very complex nor very simple:
$\Rightarrow P (E_{1}^{'} \cap E_{5}^{'}) = P (E_{1} \cup E_{5})^{'} = 1 - P (E_{1} \cup E_{5})$
$= 1 - [P (E_{1}) + P (E_{5})]$
$= 1 - [0.15 + 0.08]$
$= 1 - 0.23$
$∴ P (E_{1}^{'} \cap E_{5}^{'}) = 0.77$
(iii) To find the probabilities that a particular surgery will be rated Routine or complex:
$\Rightarrow P (E_{3} \cup E_{2}) = P (E_{3}) + P (E_{2}) - P (E_{3} \cap E_{2})$
$= 0.31 + 0.2 - 0$
$∴ P (E_{3} \cup E_{2}) = 0.51$ .
(iv) To find the probabilities that a particular surgery will be rated routine or simple:
$\Rightarrow P (E_{3} \cup E_{4}) = P (E_{3}) + P (E_{4}) - P (E_{3} \cap E_{4})$
$= 0.31 + 0.26 - 0$
$∴ P (E_{3} \cup E_{2}) = 0.57$ .

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