Question

certain virus infects one in every 300 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". Hint: Make a Tree Diagram a) Find the probability that a person has the virus given that they have tested positive, i.e. find P(A|B). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(A|B)= % b) Find the probability that a person does not have the virus given that they test negative, i.e. find P(A'|B'). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(A'|B') =

Answer 1

Answer:

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.

Step-by-step explanation:

a) The probability of B: "the person tests positive" is the sum of:

- The probability that a person has the virus and the test is positive.

$P(A) \cdot P(B|A)$

- The probability that a person don't have the virus but the test is positive.

$P(not \,A) \cdot P(B|not\,A)$

Then, the probability of B: "the person tests positive" is

$P(A|B)=\frac{P(B|A)}{P(B)}=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}$

We have

$P(A)=1/300=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B|A)=0.8\\\\P(B|A')=0.08$

Then

$P(A|B)=\frac{P(A)\cdot P(B|A)}{P(A)\cdot P(B|A)+P(A')\cdot P(B|A')}\\\\P(A|B)=\frac{0.0033*0.8}{0.0033*0.8+0.9967*0.08}= \frac{0.00264}{0.00264+0.07974}=\frac{0.00264}{0.08238}= 0.032$

The probability that a person has the virus given that they have tested positive is P(A|B)=0.032.

b) The probability P(A'|B') is

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}$

We have

$P(A)=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B'|A)=1-0.8=0.2\\\\P(B'|A')=1-0.08=0.92$

Then

$P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}=\frac{0.92*0.9967}{0.92*0.9967+0.2*0.0033}= \frac{0.917}{0.917+0.00066}= 0.999$

The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.