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
Let
E1 = event that surgeries are rated as very complex
E2 = event that surgeries are rated as complex
E3 = event that surgeries are rated as routine
E4 = event that surgeries are rated as simple
E5 = event that surgeries are rated as very simple
Given: P (E1) = 0.15, P (E2) = 0.20, P (E3) = 0.31, P (E4) = 0.26, P (E5) = 0.08
(a) P (complex or very complex) = P (E1 or E2) = P (E1⋃ E2)
By General Addition Rule:
P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
⇒ P (E1⋃ E2) = P (E1) + P (E2) – P (E1⋂ E2)
= 0.15 + 0.20 – 0 [given] [∵ All events are independent]
= 0.35
(b) P (neither very complex nor very simple) = P (E1’ ⋂ E5’)
= P (E1⋃ E5)’
= 1 – P (E1⋃ E5)
[∵By Complement Rule]
= 1 – [P (E1) + P (E5) – P (E1⋂ E5)] [∵ By General Addition Rule]
= 1 – [0.15 + 0.08 – 0]
= 1 – 0.23
= 0.77
(c) P (routine or complex) = P (E3⋃ E2)
= P (E3) + P (E2) – P (E3⋂ E2)
[∵ By General Addition Rule]
= 0.31 + 0.20 – 0 [given]
= 0.51
(d) P (routine or simple) = P (E3⋃ E4)
= P (E3) + P (E4) – P (E3⋂ E4)
[∵ By General Addition Rule]
= 0.31 + 0.26 – 0 [given]
= 0.57
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