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;
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