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  • 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 pro

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

Answers (1)

Given:

P (E1) = 0.15

P (E2) = 0.20

P (E3) = 0.31

P (E4) = 0.26

P (E5) = 0.08

Let us consider that –

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

  1. P (complex or very complex) = P (E1 or E2)

= P (E1 U E2)

Now, by the general rule –

P (A U B) = P(A) + P(B) – P(A ∩ B)

We have,                    

 P (E1 U E2) = P(E1) + P(E2) – P(E1 ∩ E2)

= 0.15 + 0.2 – 0

= 0.35

  1. P (neither very complex nor very simple) = P (E1’ ∩ E5’)

= 1 - P (E1 ∩ E5) ….. (by complement rule)

= 1 – [P (E1) + P (E5) – P (E1 ∩ E5)  … (general addn rule)

= 1 – [0.15 + 0.08 – 0]

= 1 – 0.23

= 0.77

  1. P (routine or complex) = P (E3 ∩ E2)

= P (E3) + P (E2) - P (E3 ∩ E2) …… (by general addition rule)

= 0.31 + 0.2 – 0

= 0.51

  1. P (routine or simple) = P (E3 ∩ E4)

= P (E3) + P (E4) - P (E3 ∩ E4) …… (by general addition rule)

= 0.31 + 0.26 - 0

= 0.57

Posted by

infoexpert22

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