Question

g A population is infected with a certain infectious disease. It is known that 95% of the population has not contracted the disease. A test for this disease is 98% accurate (i.e., a person who has contracted the disease tests positive 98% of the time) and has a 1% false negative rate (i.e., a person without the disease has 1% positive rate). Find the probability that a random selected person from does not have the infection if he or she has tested positive. Briefly explain why you are or are not surprised by your result.

Answer 1

Answer:

There is approximately 17% chance of a person not having a disease if he or she has tested positive.

Step-by-step explanation:

Denote the events as follows:

D = a person has contracted the disease.

+ = a person tests positive

- = a person tests negative

The information provided is:

$P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01$

Compute the missing probabilities as follows:

$P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99$

The Bayes' theorem states that the conditional probability of an event, say A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}$

Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:

$P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}$

              $=\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667$

So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.

As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.