Question

A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 8% of the time if the person does not have the virus. (This 8% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".

Answer 1

Answer:

a) $P(A|B)=\frac{10}{209}$=0.048=4.8%

(b) $P(A^c|B^c)=\frac{4577}{4582}$=0.999=99.9%

Step-by-step explanation:

Note that:

A: the person is infected

$A^c$: the person is not infected

B: the person tests positive

$B^c$: the person tests negative

a) When a person tests positive, the probability that the person is infected is given by:

P(A|B)=$\frac{\frac{1}{200} *\frac{80}{100} }{\frac{199}{200} *\frac{8}{100}+\frac{1}{200}*\frac{80}{100}} =\frac{10}{209}$

Checking on the tree diagram we can see that there are two ways in which the test is positive(the first one and the third one). These ways will be the total outcomes of the probability equation. The favorable outcome is the first one.

P(A|B)=$\frac{\frac{1}{200} *\frac{80}{100} }{\frac{199}{200} *\frac{8}{100}+\frac{1}{200}*\frac{80}{100}} =\frac{10}{209}$

b) When a person tests negative, the probability that the person is not infected is given by:

$P(A^c|B^c)$

Checking on the tree diagram we can see that there are two ways in which the test is negative(the second one and the fourth one). These ways will be the total outcomes of the probability equation. The favorable outcome is the fourth one.

$P(A^c|B^c)=\frac{\frac{199}{200} *\frac{92}{100} }{\frac{199}{200} *\frac{92}{100}+\frac{1}{200}*\frac{20}{100}} =\frac{4577}{4582}$=0.999=99.9%