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;

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) complex or very complex; (b) neither very complex nor very 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) complex or very complex;
(b) neither very complex nor very 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) complex or very complex;
(b) neither very complex nor very 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) $P$ (complex or very complex) $=P\left(E_{1}\right.$ or $\left.E_{2}\right)=P\left(E_{1} \cup E_{2}\right)$
Using the General Addition Rule:
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)$
$=0.15+0.20-0$ [as given] $[\because$ All events are independent] $=0.35$
(b) $P$ (neither very complex nor very simple) $=P\left(E_{1}^{\prime} \cap E_{5}^{\prime}\right)$
$=P\left(E_{1} \cup E_{5}\right)^{\prime}$
[:By the Complement Rule]
$=1-\left[\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{5}\right)-\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{5}\right)\right][\because$ By the General Addition Rule]
$=1-[0.15+0.08-0]$
$=1-0.23$
$=0.77$