Question

The proportion of people in a given community who have a certain disease is 0.005. A test is available to diagnose the disease. If a person has the disease, the probability that the test will produce a positive signal is 0.96. If a person does not have the disease, the probability that the test will produce a positive signal is 0.04.

1. If a man tests negative, what is the probability that he actually has the disease?

2. For many medical tests, it is standard procedure to repeat the test when a positive signal is given. If repeated tests are independent, what is the probability that a man will test positive on two successive tests if he has the disease?

3. Assuming repeated tests are independent, what is the probability that a man tests positive on two successive tests if he does not have the disease? Round the answer to four decimal places.

4. If a man tests positive on two successive tests, what is the probability that he has the disease?

Answer 1

1)P(tested negative)=P(has disease and tested negative)+P(not has disease and tested negative)

=0.005*(1-0.96)+(1-0.005)*(1-0.04)=0.9554

P(has disease given tested negative)

=P(has disease and tested negative)/P(tested negative)=0.005*(1-0.96)/0.9554=0.000209

2)

P(tested positive twice given has disease)=0.96*0.96=0.9216

3)

P((tested positive twice given not has disease)=0.04*0.04=0.0016

4)

P(tested positive twice)=P(has disease and tested positive twice )+P(not has disease and tested positive twice )=0.005*0.96*0.96+(1-0.005)*0.04*0.04=0.0062

hence P(has disease given tested positive twice)

=P(has disease and tested positive twice )/P(tested positive twice)

=0.005*0.96*0.96/0.0062=0.7432