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, $0.15$, $0.20$, $0.31$, $0.26$ and $0.08$. Find the probabilities that a particular surgery will be rated
(i) complex or very complex
(ii) neither very complex nor very simple
(iii) routine or complex
(iv) routine or simple

	(i)	(ii)	(iii)	(iv)
(a)$\,\,\,\,$	$0.77\,\,\,\,$	$0.35\,\,\,\,$	$0.51\,\,\,\,$	$0.57\,\,\,\,$
(b)	$0.35$	$0.77$	$0.51$	$0.57$
(c)	$0.35$	$0.51$	$0.77$	$0.57$
(d)	$0.57$	$0.77$	$0.51$	$0.35$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

Let $A$, $B$, $C$, $D$ and $E$ be the events that surgeries are rated as very complex, complex, routine, simple or very simple respectively.
$\therefore P(A) = 0.15$, $P(B) = 0.20$, $P(C) = 0.31$,
$P (D) = 0.26$ and $P(E) = 0.08$
(i) $P$ (complex or very complex) $= P(A$ or $B)$
$= P(A \cup B ) = P(A) + P(B) - P(A \cap B)$
$= 0.15 + 0.2 0 - 0 \quad [\because P (A \cap B ) = 0]$
$= 0.35$
(ii) $P$(neither very complex nor very simple)
$= P(A' \cap E') = P(A \cup E)'$
$= 1 - P(A \cup E) = 1 - [P(A) + P(E)] \quad[ \because P(A \cap E) = 0]$
$= 1 - ( 0 .1 5 + 0.08) = 1 - 0 .2 3 = 0.77$
(iii) $P$(routine or complex) $= P(C \cup B)$
$= P (C )+ P (B ) \quad [ \because P (C \cap B) = 0]$
$= 0 .3 1 + 0 .2 0 = 0.51$
(iv) $P$(routine or simple) $= P(C \cup D)$
$= P(C) + P(D) \quad [\because P(C \cap D) = 0]$
$= 0.31 + 0.26 = 0.57$