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Math Help - Conditional Probability

  1. #1
    uhm
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    Conditional Probability

    Hi,

    I'm a first year at Uni and I've just learnt Probability this year for the first time. I dont know if this was the right place to post this question but I'm struggling with understanding conditional probability.

    The question I've been trying to work out is asking me to find the probability of

    p(B=1|A=1)

    where

    A=1 Ill
    A=0 Not ill

    B=1 Test is positive
    B=0 Test is negative

    The test is 95% reliable
    95% of cases are reported as positive
    95% of cases are reported as negative
    1% of people within the same age range and the same background have the illness

    I'd really appreciate it if someone could just explain to me how to find the values to solve the p(B=1|A=1) problem,

    Thank you!
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  2. #2
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    \displaystyle P(A=1|B=1) = \frac{P(A=1 \cap B=1)}{P(B=1)}
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  3. #3
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    Sounds like a Bayesian problem. Is that something you're covering or have covered? It's a conditional probability formula:

    P(B\ |\ A) = \frac{P(A\ |\ B)P(B)}{P(A)}

    Let us assume A, B is 1 and not-A, B is 0. We say that the probability of B given the hypothesis A is equal to the probability of B (prior probability; prior to any other information or hypothesis) multiplied by the likelihood of A given B (i.e., if they report positive, how likely are they to be ill?), quantity divided by the normalizing constant (or prior) probability of A. Think about the information you're given and the conditional probability in terms of Bayes theorem.
    Last edited by bryangoodrich; May 23rd 2011 at 02:03 PM. Reason: latex correction
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  4. #4
    uhm
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    Hi,

    Thank you for replying,

    We didnt cover it as Bayes theorem, instead we were told that P(B|A)= P(B and A)/P(A)

    The question I have is how can I derive the values so that it would be like this for example: 0.34/0.47 =0.72

    Unless that isnt actually what the conditional probability is of course
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  5. #5
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    Quote Originally Posted by uhm View Post
    The question I've been trying to work out is asking me to find the probability of
    p(B=1|A=1)
    where
    A=1 Ill
    A=0 Not ill

    B=1 Test is positive
    B=0 Test is negative

    The test is 95% reliable
    95% of cases are reported as positive
    95% of cases are reported as negative
    1% of people within the same age range and the same background have the illness
    Surely you have garbled this question.

    READ:
    95% of cases are reported as positive
    95% of cases are reported as negative.
    What does that mean?

    What does "The test is 95% reliable" mean?

    Usually it means \mathcal{P}(A=1|B=1)=0.95 and
    \mathcal{P}(A=0|B=0)=0.95.
    Is this what you understand?
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  6. #6
    uhm
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    Hi,

    I understand it sounds a bit garbled, this was taken from an old exam question because we were never actually set any work on this during the course of being taught it. I didnt know if repeating the entire question was against the forum rules as this is my first thread.

    The test is 95% reliable, is one of the pieces of information given within the explanation of the problem, I wrote it down to see if it was actually helpful.

    And yes I under stand that both equal 0.95 but say the probability changed to A=1 and B=0 how would it change the probability, since A and B are not the same?

    Thank you for your help!
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  7. #7
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    Well Bayes' theorem is just a formulation using conditional probability. In particular, it is used for the sort of questions you're asking. Just look at the first example on that wiki article I linked. I also have to agree with Plato, after thinking about your question I had to wonder what information you provided. I think you're trying to say that the success rate of the test is 95%. Thus, the failure rate of the test is 5%. If we put these in terms of "+" and "-" as in the wikipedia example, you'll see it parallels your question in a lot of ways. This sort of success rate of some medical test is an often used example for Bayes' theorem after one understands conditional probability, because the conditional statement is included in the theorem. Also, are you sure you want P(B | A) and not P(A | B)? Just curious. I would translate P(B | A) as "the probability of a positive test given that the person is ill." Well, the answer is the success rate, since it doesn't matter if the person is ill or not. The test is 95% accurate, right? This is why I wondered if you didn't want P(A | B). This question is far more interesting: the probability that someone is ill given a positive test result. This forces us to consider false positives, etc.

    Actually, not everything I said above was correct. P(B | A) is correct 95% of the time. If we use the notation Plato used above, that is precisely what it indicates: i.e., P(B = 1 | A = 1) = P(B = 0 | A = 0) = 0.95.

    I think it would be good to start listing out the probabilities that we know, and then figure out any conditional probabilities. Once you've translated the known information into something useful, we are in a more apt position to make use of it. So what does our information tell us?
    Last edited by bryangoodrich; May 23rd 2011 at 03:07 PM. Reason: thought correction
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  8. #8
    uhm
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    The probability in the question was P(B=1|A=1) which did confuse me with the information given. I will have a look at Bayes theorem and see if I can work it out with that,

    Thank you for helping! I hope it doesn't mean there could be false positives!
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  9. #9
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    Well as was already pointed out, the test is 95% reliable. So you expect a positive test when the person is ill 95% of the time. Thus, the answer is given immediately from your information.

    P(B = 1\ |\ A = 1) = 0.95

    I would recommend for practice that you figure out some of the other probabilities (e.g, P(A = 0), P(A = 1), P(A = 1 | B = 1), etc.). Check out the wikipedia article example 1 I linked. The structure of its problem is identical (if we think of B as "+", A = 1 as "D" or "diseased", and A = 0 as "N" or "Not Sick"). Work out the probabilities to get a better understanding. This example doesn't really show you much of anything about calculating conditional probabilities!
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  10. #10
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    Quote Originally Posted by uhm View Post
    Hi,

    I understand it sounds a bit garbled, this was taken from an old exam question because we were never actually set any work on this during the course of being taught it. I didnt know if repeating the entire question was against the forum rules as this is my first thread.

    The test is 95% reliable, is one of the pieces of information given within the explanation of the problem, I wrote it down to see if it was actually helpful.

    And yes I under stand that both equal 0.95 but say the probability changed to A=1 and B=0 how would it change the probability, since A and B are not the same?
    Well, ok. Can't say for sure that I follow you.
    But it seems that you want \mathcal{P}(B=1|A=1)=\frac{\mathcal{P}(A=1|B=1){P}  (B=1)}{\mathcal{P}(A=1)}
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