Question

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 Select the correct answer from above options

Answer 1

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