A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive 90% 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".
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') =
Given probabilities:
P(A) = 1/200 = 0.005
P(B) = 199/200 = 0.995
P(B|A) = 0.9
P(B|~A) = 0.1
Infer:
P(~B|A) = 0.1
P(~B|~A) = 0.9
Then:
a.) Find P(A|B)
P(A|B) = P(A)P(B|A) = 0.005×0.9 = 0.0045 rounds to 0.0
b.) Find P(~A|~B)
P(~A|~B) = P(~A)P(~B|~A) = 0.995×0.9 = 0.896 rounds to 0.9
Note: the symbol ~A means A' in your notation.
QED
To find the probabilities in this scenario, we can use Bayes' theorem. Bayes' theorem states that:
P(A|B) = (P(B|A) * P(A)) / P(B)
a) To find P(A|B), we need to find P(B|A), P(A), and P(B).
First, let's find P(B|A), which is the probability that a person tests positive given that they have the virus. We are given that the test is positive 90% of the time if the person has the virus, so P(B|A) = 0.9.
Next, we need to find P(A), which is the probability that a person has the virus. We are told that the virus infects 1 in every 200 people, so P(A) = 1/200 = 0.005.
Finally, we need to find P(B), which is the probability that a person tests positive. This can be calculated using the law of total probability.
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
We know that P(B|A) = 0.9, and P(B|A') is the probability of a false positive, which is 10%. Since 1 person in 200 has the virus, the probability of not having the virus (A') is 199/200. Therefore, P(A') = 199/200 = 0.995. The probability of a false positive is 10% or 0.10.
P(B) = 0.9 * 0.005 + 0.10 * 0.995
P(B) = 0.0045 + 0.0995
P(B) = 0.104
Now we can substitute these values into Bayes' theorem to find P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (0.9 * 0.005) / 0.104
P(A|B) = 0.0045 / 0.104
P(A|B) = 0.0433
Therefore, the probability that a person has the virus given that they have tested positive is approximately 0.0433, or 4.3%.
b) To find P(A'|B'), we need to find P(B'|A'), P(A'), and P(B').
P(B'|A') is the probability that a person tests negative given that they do not have the virus. We are told that the test is positive 10% of the time if the person does not have the virus, so P(B'|A') = 0.1.
P(A') is the probability that a person does not have the virus. We already calculated this in part a), and found that P(A') = 0.995.
P(B') is the probability that a person tests negative. This can also be calculated using the law of total probability.
P(B') = P(B'|A) * P(A) + P(B'|A') * P(A')
We know that P(B'|A') = 0.1, and P(B'|A) is the probability of a false negative, which is 10% or 0.10.
P(B') = 0.1 * 0.005 + 0.10 * 0.995
P(B') = 0.0005 + 0.0995
P(B') = 0.1
Now we can substitute these values into Bayes' theorem to find P(A'|B'):
P(A'|B') = (P(B'|A') * P(A')) / P(B')
P(A'|B') = (0.1 * 0.995) / 0.1
P(A'|B') = 0.0995 / 0.1
P(A'|B') = 0.995
Therefore, the probability that a person does not have the virus given that they test negative is approximately 0.995, or 99.5%.
REALLY BIG HINT. Draw the probability tree. The answers come quickly with that.
Read through the nearly identical posts linked in Related Questions below. I don't know if any were answered, but it's worth looking.
And if you'd follow directions and put your SUBJECT in the SUBJECT box, that would really help.