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 $B)$
$=P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=0.15+0.20-0[\because P(A\cap B)=0]$
$=0.35$
(ii) $P$(neither very complex nor very simple)
$=P(A^{\prime }\cap E^{\prime })=P(A\cup E{)}^{\prime }$
$=1-P(A\cup E)=1-[P(A)+P(E)][\because P(A\cap E)=0]$
$=1-(0.15+0.08)=1-0.23=0.77$
(iii) $P$(routine or complex) $=P(C\cup B)$
$=P(C)+P(B)[\because P(C\cap B)=0]$
$=0.31+0.20=0.51$
(iv) $P$(routine or simple) $=P(C\cup D)$
$=P(C)+P(D)[\because P(C\cap D)=0]$
$=0.31+0.26=0.57$