Question

A certain virus infects 5% of the population. A test used to detect the virus in a person is positive 80% of the time if the person has the virus, and 10% of the time if the person does not have the virus.

a. What is the probability that a randomly selected person tested positive and has the virus?

b. What is the probability that a randomly selected person tested positive and does not have the virus?

c. What is the probability that a randomly selected person has the virus given that they get a positive test result?

Show which rule you use to calculate the probability!

Answer 1

D+→having virus.(inflected?

D-→not having virus(not inflected)

T+→test is positive

P(D+)= 5%=0.05

P(D-)= 0.95

P(T+/D+)= 80%=0.80

P(T+/D-)= 0.10

Now we have to find

a)

the probability that a randomly selected person tested positive and has the virus

=P(T+/D+)

=0.80

b)

the probability that a randomly selected person tested positive and does not have the virus

= P(T+/D-)

= 0.10

C)#using bays theorem:

the probability that a randomly selected person has the virus given that they get a positive test result

P(D+/T+)

= [P(D+)*P(T+/D+)]÷[P(D+)*P((T+/D+)+P(D-)*P(T+/D-)]

=(0.05*0.80)/(0.05*0.80+0.95*0.10)

=0.296296

Thanks!

Stay safe!