sometimes just using numbers will clarify
for 2000 people
... 10 have the virus
... 8 will test pos ... 2 will not
... 1990 do not have the virus
... 199 will test pos
a) 207 people test pos
... 8 of them have it
... 8/207 = .0386
b) 1793 people test neg
... 2 of them have it
... 1791/1793 = .9989
I am enrolled in an online Math 125 class. I posted this question earlier in the week and never got a response, so here I am! I am just not understanding these questions. I made the tree diagram as suggested and put all the information at the bottom how I worked the problem out. It is just super confusing, and I can't seem to find anything that helps me work this problem out. Any help would be appreciated! Thanks!
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 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".
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.
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.
.80 Positive Test Result +
.005 Infected
.20 Negative Test Result --
Tested
.10 Positive Test Result +
.995 Not Infected
.90 Negative Test Result --
I believe the answer that I got for part (a) was 3.8%. I didn't try part (b) since I couldn't find the correct answer for part (a).
2 answers
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 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".
P(infected) = 1/200
P(not infected) = 199/200
--------------------------------
P(A) = 1/200 ; P(A') = 199/200
-------
P(B|A') = 0.1
P(B|A) = 0.8 ; P(B'|A) = 0.2
P(B|A) = P(test positive | infected) = 0.80
P(A'|B) = P(is not infected | test positive) = 0.10
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) = P(B and A)/P(B) = P(B|A)*P(A)/P(B)
= P(B|A)P(A)/[P(B and A)+P(B and A')
= P(B|A)P(A)/[P(B|A)P(A)+P(B|A')P(A')]
= [0.8*(1/200)]/[0.8*(1/200) + 0.1*(199/200)]
= 0.004/[0.004+ 0.104]
= 0.037
------------------
Cheers,
P(infected) = 1/200
P(not infected) = 199/200
--------------------------------
P(A) = 1/200 ; P(A') = 199/200
-------
P(B|A') = 0.1
P(B|A) = 0.8 ; P(B'|A) = 0.2
P(B|A) = P(test positive | infected) = 0.80
P(A'|B) = P(is not infected | test positive) = 0.10
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) = P(B and A)/P(B) = P(B|A)*P(A)/P(B)
= P(B|A)P(A)/[P(B and A)+P(B and A')
= P(B|A)P(A)/[P(B|A)P(A)+P(B|A')P(A')]
= [0.8*(1/200)]/[0.8*(1/200) + 0.1*(199/200)]
= 0.004/[0.004+ 0.104]
= 0.037
------------------
Cheers,
1 answer