A team of medical students doing their internship

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

Solution:

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

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now