**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, .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**

LetE

_{1}= event that surgeries are rated as very complexE

_{2}= event that surgeries are rated as complexE

_{3}= event that surgeries are rated as routineE

_{4}= event that surgeries are rated as simpleE

_{5}= event that surgeries are rated as very simpleGiven: P (E_{1}) = 0.15, P (E_{2}) = 0.20, P (E_{3}) = 0.31, P (E_{4}) = 0.26, P (E_{5}) = 0.08(a) P (complex or very complex) = P (E_{1}or E_{2}) = P (E_{1}⋃ E_{2})By General Addition Rule:

P (A ∪ B) = P(A) + P(B) – P (A ∩ B)

⇒ P (E

_{1}⋃ E_{2}) = P (E_{1}) + P (E_{2}) – P (E_{1}⋂ E_{2})= 0.15 + 0.20 – 0 [given] [∵ All events are independent]

= 0.35

(b) P (neither very complex nor very simple) = P (E_{1}’ ⋂ E_{5}’)= P (E

_{1}⋃ E_{5})’= 1 – P (E

_{1}⋃ E_{5})[∵By Complement Rule]

= 1 – [P (E

_{1}) + P (E_{5}) – P (E_{1}⋂ E_{5})] [∵ By General Addition Rule]= 1 – [0.15 + 0.08 – 0]

= 1 – 0.23

= 0.77

(c) P (routine or complex) = P (E_{3}⋃ E_{2})= P (E

_{3}) + P (E_{2}) – P (E_{3}⋂ E_{2})[∵ By General Addition Rule]

= 0.31 + 0.20 – 0 [given]

= 0.51

(d) P (routine or simple) = P (E_{3}⋃ E_{4})= P (E

_{3}) + P (E_{4}) – P (E_{3}⋂ E_{4})[∵ By General Addition Rule]

= 0.31 + 0.26 – 0 [given]

= 0.57