Diagnosing HIV: Assessing Probability with a Test

[chrisyuen]

Homework Statement

A diagnostic test is used to detect the HIV virus. It is known that 2% of the people in a city have the virus. Extensive research on the diagnostic test reveals that its results are correct 95% of the time. In other words, whether or not an individual has the HIV virus, the probability that the test gives a correct diagnosis is 0.95.

(a) A person took the test and was diagnosed to have the HIV virus. What is the probability that he actually did not have the virus?

(b) Three persons took the test and were all diagnosed to have the HIV virus. What is the probability that at least two of them have the virus.

(Answers:
(a) 0.7206
(b) 0.1906)

Homework Equations

Probability Formulae

The Attempt at a Solution

(a) 0.98 x 0.05 / (0.98 x 0.05 + 0.02 x 0.95) = 0.720588
(b) 1 - (0.7206)³ - 3 x (0.7206)² x (1 - 0.7206) = 0.19057

For part (a), I don't know why the answer can be calculated from the above formula.

Can anyone explain it to me?

Tree Diagram:

B1 - HIV
B1.1 - Y
B1.2 - N
B2 - No HIV
B2.1 - Y
B2.2 - N

B : Branch; Y : Correct Diagnosis, N : Incorrect Diagnosis

In addition, may the above formula conflict with the tree diagram I drew?

Thank you very much!

 

[lanedance]

Hi chrisyuen

part a) looks like conditional probability to me, do you know the conditional probability formula?

i think your tree looks ok and the the branches match the prob you are given

if you write in the probabilities, your tree contains all the info for the conditional probability formula to get a)

 

[HallsofIvy]

Homework Statement

A diagnostic test is used to detect the HIV virus. It is known that 2% of the people in a city have the virus. Extensive research on the diagnostic test reveals that its results are correct 95% of the time. In other words, whether or not an individual has the HIV virus, the probability that the test gives a correct diagnosis is 0.95.

(a) A person took the test and was diagnosed to have the HIV virus. What is the probability that he actually did not have the virus?

Imagine that 100000 people take the test. Since 2% of people have the virus, 2000 have the virus and 98000 do not. Of the 2000 who have the virus, 0.95(2000)= 1900 test positive. Of the 98000 who do not, 0.05(98000)= 4900 also test positive. That is, of 1900+ 4900= 6800 people who test positive, 1900 actually have the virus. 1900/6800= 0.28.

(b) Three persons took the test and were all diagnosed to have the HIV virus. What is the probability that at least two of them have the virus.

(Answers:
(a) 0.7206
(b) 0.1906)

Homework Equations

Probability Formulae

The Attempt at a Solution

(a) 0.98 x 0.05 / (0.98 x 0.05 + 0.02 x 0.95) = 0.720588
(b) 1 - (0.7206)³ - 3 x (0.7206)² x (1 - 0.7206) = 0.19057

For part (a), I don't know why the answer can be calculated from the above formula.

It looks to me like it clearly can't be.

Can anyone explain it to me?

Tree Diagram:

B1 - HIV
B1.1 - Y
B1.2 - N
B2 - No HIV
B2.1 - Y
B2.2 - N

B : Branch; Y : Correct Diagnosis, N : Incorrect Diagnosis

In addition, may the above formula conflict with the tree diagram I drew?

Thank you very much!

 

[chrisyuen]

Thanks lanedance & HallsofIvy!

After I read the example from HallsofIvy, I knew that I misunderstood the question before and now understand it deeply.

Thank you very much!