A team of medical students doing their internship has 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:

(a) complex or very complex;

(b) neither very complex nor very simple;

(c) routine or complex;

(d) routine or simple.

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Payal Gupta | Contributor-Level 10

8 months ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t E_{1}, E_{2}, E_{3}, E_{4} a n d E_{5} b e t h e e v e n t s t h a t t h e s u r g e r i e s a r e r a t e d a s v e r y c o m p l e x, c o m p l e x, \\ r o u t i n e, s i m p l e a n d v e r y s i m p l e r e s p e c t i v e l y . \\ ∴ P (E_{1}) = 0.15, P (E_{2}) = 0.20, P (E_{3}) = 0.31, P (E_{4}) = 0.26 a n d P (E_{5}) = 0.08 \\ (a) P (c o m p l e x a n d v e r y c o m p l e x) = P (E_{1} o r E_{2}) \\ \Rightarrow P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) \\ = 0.15 + 0.20 - 0 \\ = 0.35 [? A l l e v e n t a r e i n d e p e n d e n t] \\ (b) P (n e i t h e r v e r y c o m p l e x n o r v e r y s i m p l e) = P (E_{1}^{'} \cap E_{5}^{'}) \\ = P (E_{1} \cup E_{5})' = 1 - P (E_{1} \cup E_{5}) \\ = 1 - [P (E_{1}) + P (E_{5})] = 1 - [0.15 + 0.08] \\ = 1 - 0.23 = 0.77 \\ (c) P (r o u t i n e o r c o m p l e x) = P (E_{3} o r E_{2}) \\ \Rightarrow P (E_{3} \cup E_{2}) = P (E_{3}) + P (E_{2}) \\ = 0.31 + 0.20 = 0.51 \\ (d) P (r o u t i n e o r s i m p l e) = P (E_{3} o r E_{4}) \\ \Rightarrow P (E_{3} \cup E_{4}) = P (E_{3}) + P (E_{4}) \\ = 0.31 + 0.26 = 0.57 \end{array}$

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