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Math Help - Conditional probability

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    Exclamation Conditional probability

    I've been stuck on this and can not for the life of me figure it out any help would be appreciated.

    A test is 97% accurate that a person who tests positive actually has the disease.If 2% of the population has the disease, what is the probability that a person selected at randon actually has the disease if they test positive?
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    Quote Originally Posted by nightrider456 View Post
    I've been stuck on this and can not for the life of me figure it out any help would be appreciated.

    A test is 97% accurate that a person who tests positive actually has the disease.If 2% of the population has the disease, what is the probability that a person selected at randon actually has the disease if they test positive?
    This seems to me a trick question. Usually for this type of question the wording is different, and there's calculation you need to do.

    But here we are given the probability that's being asked for. That is, the answer is simply 97%.

    The information about choosing a person at random and 2% of the population has the disease is irrelevant. Unless maybe you copied it down wrong?
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    Quote Originally Posted by undefined View Post
    This seems to me a trick question. Usually for this type of question the wording is different, and there's calculation you need to do.

    But here we are given the probability that's being asked for. That is, the answer is simply 97%.

    The information about choosing a person at random and 2% of the population has the disease is irrelevant. Unless maybe you copied it down wrong?
    I double checked it and it's word for word the correct answer is 0.3975
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  4. #4
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    Quote Originally Posted by nightrider456 View Post
    I double checked it and it's word for word the correct answer is 0.3975
    I see now that I didn't think carefully enough. What convinced me was considering the scenario of a population that is 100% diseased, and then a random subject is chosen and the test comes out positive, obviously the answer in that case would not be 97%. Sorry for being too hasty and posting an incorrect response.

    My new answer still may not be helpful to you, but I think it's correct.

    Now I believe there's not enough information. The reasoning is as follows.

    Let p_1 be the probability that a diseased person tests positive.

    Let p_2 be the probability that a healthy person tests positive.

    The number 97% isn't as clearly defined as I would like, but it seems to mean: choose a subject A, with 50% probability that A is diseased and 50% probability A is healthy. Then 97% represents the probability that A is diseased if the test comes out positive.

    In terms of p_1 and p_2, we can write

    \frac{0.5p_1}{0.5p_1+0.5p_2}=0.97

    which I got using a decision tree, but with the same reasoning described in this example on Wikipedia.

    So we don't know p_1 and p_2, only how they relate to each other.

    When we solve the problem being asked, we find the desired probability equal to

    p=\frac{0.02p_1}{0.02p_1+0.98p_2}

    So there are two equations and three unknowns, and there is not enough information to solve.
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    Quote Originally Posted by nightrider456 View Post
    I've been stuck on this and can not for the life of me figure it out any help would be appreciated.

    A test is 97% accurate that a person who tests positive actually has the disease.If 2% of the population has the disease, what is the probability that a person selected at randon actually has the disease if they test positive?
    97% who test positive have the disease.
    3% who test positive do not have the disease.

    2% of the population test positive at a 97% accuracy.
    98% of the population test positive at a 3% innacuracy (incorrectly).

    Hence the probability that a positive test indicates disease is

    \frac{(0.02)(0.97)}{(0.02)(0.97)+(0.03)(0.98)}=\fr  ac{0.0194}{0.0488}=\frac{194}{488}=0.39754
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    97% who test positive have the disease.
    3% who test positive do not have the disease.

    2% of the population test positive at a 97% accuracy.
    98% of the population test positive at a 3% innacuracy (incorrectly).
    In an attempt to prove that this method involves a hidden assumption, I have verified your result.

    I proceeded as follows:

    Suppose the probability of a false negative is 0.5, and the probability of a false positive is 3/194 \approx 0.0154639. Then suppose we have a room with two people in it: one diseased and one healthy. We choose a person at random and perform the test. The test is positive. We can verify that the probability that the person is the diseased one is exactly 0.97.

    I expected the final answer to be different, but it was the same answer you gave. So then I worked it out:

    \frac{0.5p_1}{0.5p_1+0.5p_2}=0.97

    \left(\frac{p_1}{2}+\frac{p_2}{2}\right)\cdot 0.97=\frac{p_1}{2}

    (p_1+p_2)\cdot 0.97 = p_1

    p_2 = \left(\frac{3}{97}\right)p_1

    And then the desired probability:

    p=\frac{0.02p_1}{0.02p_1+0.98p_2}

    p=\frac{0.02p_1}{0.02p_1+0.98\cdot(3/97)\cdot p_1}

    p=\frac{0.02}{0.02+0.98\cdot(3/97)}

    \approx 0.39754

    So I stopped too soon when I concluded that there was not enough information. Ah well, I know for next time. Sorry again for an incorrect response.
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