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 event 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