Question

A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 85% of the time if the person has the virus and 5% of the time if the person does not have the virus. (This 5% result is called a false positive.) a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest tenth of a percent and do not include a percent sign.

Answer 1

Answer: Our required probability is 0.07 i.e. 0.1.

Step-by-step explanation:

Let A be the event that "the person is infected".

Let B be the event that "the person tests positive".

Probability of person is positive if the person has the virus = 85%

Probability of person is positive if the person does not have the virus = 5%

P(B|A) = 0.85

So, P(A) = $\dfrac{1}{200}=0.005$

P(B) is given by

$\dfrac{1}{200}\times 0.85+\dfrac{199}{200}\times 0.05\\\\=0.00425+0.04975\\\\=0.054$

So, We need to find P(A|B) which is given by

$P(A|B)=P(B|A)\times \dfrac{P(A)}{P(B)}\\\\P(A|B)=0.85\times \dfrac{0.005}{0.054}\\\\P(A|B)=0.0787$

Hence, our required probability is 0.07 i.e. 0.1.