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) routine or complex (b) routine or simple

Home » 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, . Find the probabilities that a particular surgery will be rated. (a) routine or complex (b) routine or simple

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) routine or complex
(b) routine or simple

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

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) routine or complex
(b) routine or simple

Solution:
Let’ say
Event that surgeries are rated as very complex $=E_{1}$
Event that surgeries are rated as complex $=\mathrm{E}_{2}$
Event that surgeries are rated as routine $=\mathrm{E}_{3}$
Event that surgeries are rated as simple $=\mathrm{E}_{4}$
Event that surgeries are rated as very simple $=E_{5}$
Given that $P\left(E_{1}\right)=0.15, P\left(E_{2}\right)=0.20, P\left(E_{3}\right)=0.31, P\left(E_{4}\right)=0.26, P\left(E_{5}\right)=0.08$
(a) $P$ (routine or complex) $=P\left(E_{3} \cup E_{2}\right)$
$=P\left(E_{3}\right)+P\left(E_{2}\right)-P\left(E_{3} \cap E_{2}\right)$
$[\because$ By General Addition Rule]
$=0.31+0.20-0$ [given $]$
$=0.51$
(b) $P$ (routine or simple) $=P\left(E_{3} \cup E_{4}\right)$
$=P\left(E_{3}\right)+P\left(E_{4}\right)-P\left(E_{3} \cap E_{4}\right)$
$[\because$ By the General Addition Rule]
$=0.31+0.26-0$ [as given $]$
$=0.57$