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.
[tex]P(A) \cdot P(B|A)[/tex]
- The probability that a person don't have the virus but the test is positive.
[tex]P(not \,A) \cdot P(B|not\,A)[/tex]
Then, the probability of B: "the person tests positive" is
[tex]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')}[/tex]
We have
[tex]P(A)=1/300=0.0033\\\\P(A')=1-P(A)=0.9967\\\\P(B|A)=0.8\\\\P(B|A')=0.08[/tex]
Then
[tex]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[/tex]
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
[tex]P(A'|B')=\frac{P(B'|A')\cdot P(A')}{P(B'|A')\cdot P(A')+P(B'|A)\cdot P(A)}[/tex]
We have
[tex]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[/tex]
Then
[tex]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[/tex]
The probability that a person does not have the virus given that they have tested negative is P(A'|B')=0.999.
