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Math Help - Probability to get an orphan disease

  1. #1
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    Probability to get an orphan disease

    Hello all,

    p : Probability of population with a specific orphan disease (p=10^-7)
    q : Probability to get one allele called L1 on his X chromosome (q=0.01)

    The only 3 known cases of this orphan desease occured with people having the L1 allele on their X chromosome.

    What is the probability for someone to have this orphan disease knowing he has the L1 allele (with an associated confidence interval)?
    Last edited by fixisnotusedyet; December 17th 2012 at 04:58 PM.
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  2. #2
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    Re: Probability to get an orphan disease

    Hey fixisnotusedyet.

    Hint: to start you off can you express the probability in terms of a conditional statement (in other words can you define the probability as P(A|B) where you give the definitions of A and B)?

    This is half the battle in mathematics and if you do this you'll probably the solve the problem all by yourself.
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  3. #3
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    Re: Probability to get an orphan disease

    Thank you chiro.
    A=i am sick
    B=i have allele L1
    P(A)=10^-7
    P(B)=10^-2

    We are trying to calculate P(A/B)

    P(A/B)=P(A and B)/P(B)=10^2.P(A and B)

    But I have no idea how to calculate P(A and B) based on this context.
    Based on my sample of only 3 cases, observed P(A and B)=10^-7

    I am 100% sure that P(A/B) is somewhere between 0 and 10^-5.
    But giving you an answer with an 80% confidence interval... no idea if this is possible

    I know the sample is very poor (n=3) but probabilities p and q are so small that I am convinced we can still get something out of this.
    Last edited by fixisnotusedyet; December 17th 2012 at 04:48 PM.
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    Re: Probability to get an orphan disease

    Well since P(A|B) = P(A and B)/P(B) it would make sense that P(A|B) = 10^2 * 10^(-7) = 10^(-5).

    If you want to do inference, can you specify the attribute you wish to do inference on? Is it sickness or having a particular allele?
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  5. #5
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    Re: Probability to get an orphan disease

    Quote Originally Posted by fixisnotusedyet View Post
    Thank you chiro.
    A=i am sick
    B=i have allele L1
    P(A)=10^-7
    P(B)=10^-2
    We are trying to calculate P(A/B)
    P(A/B)=P(A and B)/P(B)=10^2.P(A and B)
    But I have no idea how to calculate P(A and B) based on this context.
    Based on my sample of only 3 cases, observed P(A and B)=10^-7
    I am 100% sure that P(A/B) is somewhere between 0 and 10^-5.
    But giving you an answer with an 80% confidence interval... no idea if this is possible
    I know the sample is very poor (n=3) but probabilities p and q are so small that I am convinced we can still get something out of this.

    "I know the sample is very poor (n=3)" but surely that is the whole point.
    It a heck of a difference if the population is 10^3\text{ or }10^{6}.
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  6. #6
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    Re: Probability to get an orphan disease

    As Plato hinted, P(A and B) =10^-7 is not true. This is only the assumption based on the 3 known cases...
    This is the problem with orphan disease. We do not have a population of 10^6 to confirm the assumption !

    Any idea how we can transform this false statement P(A and B) =10^-7 into a "With a probability of 80%, P(A and B) is between x and y" ?

    To clarify, the objective of my exercice is to complete the following sentence "With a probability of 80%, the probability to be sick when we have the allele L1 is between ... and ...".
    Last edited by fixisnotusedyet; December 17th 2012 at 05:04 PM.
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  7. #7
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    Re: Probability to get an orphan disease

    Hi i may have found the solution.

    T=proba of being sick having L1 allele
    t=proba of being sick.
    I calculated probabilities of having 3 cases with L1 allele for every value of T between 0 and 10^-5.
    Then I calculated the density of probability:

    With a confidence of 50% : T > 80t
    With a confidence of 80%: T>59t
    With a confidence of 95%: T>39t
    With a confidence of 99%: T>25t
    With a confidence of 99,9%: T>6t
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  8. #8
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    Re: Probability to get an orphan disease

    I reviewed what i said, it is probably wrong.

    t Probability to be sick
    p Probability to have a l1 allele
    T P(t/p)=Probability to be sick knowing we have a l1 allele
    f(T) Probability to have 3 people with a l1 allele


    For every value of T, i have calculated f(T)=(T/t*p)^3

    Si T=0.00001, f(T)=100%
    Si T=0.0000099, f(T)=97.03%
    0.0000098 94.12%
    0.0000097 91.27%
    0.0000096 88.47%
    0.0000095 85.74%
    0.0000094 83.06%
    0.0000093 80.44%
    0.0000092 77.87%
    0.0000091 75.36%
    ...
    0.0000006 0.02%
    0.0000005 0.01%
    0.0000004 0.01%
    0.0000003 0.00%
    0.0000002 0.00%
    0.0000001 0.00%
    0 0.00%

    I am able to affirm stuff like:
    If T=100t, probability to have 3/3 cases with an l1 allele is 100%.
    If T=21t, probability to have 3/3 cases with an l1 allele is 0.93%.

    (As a reminder T=100t means it is 100 times more like likely to be sick when you have a l1 allele)


    Any idea how to pursue the analysis?
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  9. #9
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    Re: Probability to get an orphan disease

    What exactly are you making an inference on? (I apologize if you already stated this, but I want to just double check).
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