What is the probability of these events?

Solution 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

Solution 2:

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 $p(\bar{A}|\bar{P}) = 1 − p(A|\bar{P}) = 1 − 0.001 = 0.999$.