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).

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

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,

To solve this problem, we can use Bayes' theorem, which relates conditional probabilities. Bayes' theorem states that:

P(A|B) = (P(B|A) * P(A)) / P(B),

where P(A|B) is the probability of event A given that event B has occurred, P(B|A) is the probability of event B given that event A has occurred, P(A) is the probability of event A, and P(B) is the probability of event B.

For part (a), we need to find P(A|B), which is the probability that a person has the virus given that they have tested positive.

From the information given in the problem, we know:
P(A) = 1/200 (one in every 200 people are infected)
P(B|A) = 0.80 (the probability of testing positive if the person has the virus)
P(B) = ? (the probability of testing positive)

To find P(B), we need to consider both cases: a person tests positive and has the virus, and a person tests positive but does not have the virus. Since the question does not provide this information directly, we can calculate it using the law of total probability.

P(B) = P(B|A) * P(A) + P(B|A') * P(A'),

where P(A') is the probability of not having the virus (1 - P(A)), and P(B|A') is the probability of testing positive if the person does not have the virus (0.10).

Substituting the known values into the equation:
P(B) = (0.80 * 1/200) + (0.10 * (1 - 1/200)) = 0.004 + 0.0995 = 0.1035

Now, we can substitute the values into Bayes' theorem to find P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B) = (0.80 * 1/200) / 0.1035 = 0.004 / 0.1035 ≈ 0.0386

So, the probability that a person has the virus given that they have tested positive is approximately 0.0386 or 3.9% (rounded to the nearest tenth of a percent). This confirms that your answer for part (a) is indeed 3.8%, which is close to the correct answer.

For part (b), we need to find P(A'|B'), which is the probability that a person does not have the virus given that they have tested negative.

Using the complementary rule, we know that:
P(A'|B') = 1 - P(A|B'),

where P(A|B') is the probability that a person has the virus given that they have tested negative.

From part (a), we found that P(A|B) is 0.0386. Therefore:
P(A'|B') = 1 - 0.0386 ≈ 0.9614

So, the probability that a person does not have the virus given that they have tested negative is approximately 0.9614 or 96.1% (rounded to the nearest tenth of a percent).

To summarize:
a) The probability that a person has the virus given that they have tested positive is approximately 3.9%.
b) The probability that a person does not have the virus given that they have tested negative is approximately 96.1%.