Answer
a) 0.99948
b) 0.32423
Work Step by Step
a) Let us consider the following events:
F=A person uses opium.
E=A person is tested positive for opium.
Observe that $P(E|F^C)=0.02$ and $P(E^C|F)=0.05$. Using Bayes' theorem,
$P(F^C|E^C)=\displaystyle \frac{P(E^C|F^C).P(F^C)}{P(E^C|F^C).P(F^C)+P(E^C|F).P(F)}$
$=\displaystyle\frac{(0.98).(0.99)}{(0.98)(0.99)+(0.05).(0.01)}=0.99948$
b)
$P(F|E)=\displaystyle \frac{P(E|F).P(F)}{P(E|F).P(F)+P(E|F^C).P(F^C)}$
$=\displaystyle\frac{(0.95).(0.01)}{(0.95)(0.01)+(0.02).(0.99)}=0.32423$
