[Math] the probability of these events

Question

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!

Answer 1

• $I$ 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