What is the probability of these events? Suppose that $4$% of the patients tested in a clinic are infected
with avian influenza. Furthermore, suppose that
when a blood test for avian influenza is given, $97$% of the
patients infected with avian influenza test positive and
that $2$% of the patients not infected with avian influenza
test positive. What is the probability that
a) a patient testing positive for avian influenza with this
test is infected with it?
b) a patient testing positive for avian influenza with this
test is not infected with it?
c) a patient testing negative for avian influenza with this
test is infected with it?
d) a patient testing negative for avian influenza with this
test is not infected with it?
Specifically, once I've found (a), is (b) just 1-(a)?  and once I've found (c), is (d) just 1-(c)?  Thank you!
 A: *

*$I$ is the event that a patient is infected, $I'$ otherwise

*$T$ is the event the patient tested positive, $T'$ otherwise

*$P(I) = 0.04$

*$P(T\,|\,I) = 0.97$

*$P(T\,|\,I') = 0.97$


a) Find $P(I\,|\,T)$
b) Find $P(I'\,|\,T) = 1-P(I\,|\,T)$
c) Find $P(I\,|\,T')$
d) Find $P(I'\,|\,T') = 1-P(I\,|\,T')$
The hint, as given is 
$P(T) \\= P(T\cap I) + P(T \cap I') \\= P(T|I) \times P(I)+ P(T|I') \times P(I')$
and that formula above is a combination of Bayes & Law of Total Probability
A: Hint: Bayes' rule
Let $Y$ be the event that a patient is infected with avian influenza.
Let $Z$ be the event that the test for avian influenza is positive.
Can you go on from here?
solution
Let A be the event that a randomly chosen person in the clinic is infected with avian influenza.
$p(A) = 0.04$  and therefore
$p(\bar{A}) = 0.96$
Let P be the event that a randomly chosen person tests positive for avian influenza on the blood test.
We are told that :
$p(P | A) = 0.97$ and $p(P | \bar{A}) = 0.02$ (“false positive”).
From these we can conclude that:
$p(\bar{P} | A) = 0.03$ (“false negative”) and $p(\bar{P} | \bar{A}) = 0.98$
a) We are asked for $p(A | P)$.
We use Bayes’ theorem:
$$P(A|P)=\frac{P(A).P(P|A)}{P(A).P(P|A)+P(\bar{A}).P(B|\bar{A})}$$
$= \frac{(0.97)(0.04)} {(0.97)(0.04) + (0.02)(0.96)}=0.669$
b) In part (a) we found $p(A | P)$.
Here we are asked for the probability of the complementary event (given a positive test result). Therefore we have simply
$p(\bar{A} | P) = 1 − p(A | P) = 1 − 0.669 = 0.331$
c) We are asked for $p(A | \bar{P})$.
We use Bayes’ theorem:
$$ p(A|\bar{P}) = \frac{p(\bar{P}|A).p(A)}{p(\bar{P}|A).p(A) + p(\bar{P}| \bar{A}).p(\bar{A})}$$
$ = \frac{(0.03)(0.04)}{(0.03)(0.04) + (0.98)(0.96)} = 0.001 $
d) In part (c) we found $p(A | \bar{P})$
Here we are asked for the probability of the complementary event (given a negative test result).
Therefore we have simply $p(\bar{A}|\bar{P}) = 1 − p(A|\bar{P}) = 1 − 0.001 = 0.999$.
