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  1. MHB Master
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    #1
    Hey!!

    The percentage of people that have a disease A is $0,01$.
    We apply twice a test for that disease, each of which give the correct answer with probability $0,95$.
    What is the probability that someone has that disease if at least one test is positive and what is the probability if both tests are positive?


    I have done the following:

    We have that $$P(\text{at least one positive})=1-P(\text{no positive})=1-P(NN)$$ where N: "negative".


    Let D be the event "has the disease".

    Then we have that $$P(NN)=P(NN\cap D)+P(NN\cap D^C)=P(NN\mid D)\cdot P(D)+P(NN\mid D^C)\cdot P(D^C)$$ We have that $P(D)=0,01$ and $P(D^C)=0,99$.

    Does it hold that $P(NN\mid D)=P(N\mid D)^2$ and $P(NN\mid D^C)=P(N\mid D^C)^2$ ?

    If yes, does it hold that $P(N\mid D)=0,05$ and $P(N\mid D^C)=0,95$ ?

    That would mean that we get $$P(NN)=0,05^2\cdot 001+0,95^2\cdot 0,99$$ Is this correct?

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  3. MHB Craftsman
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    #2
    Quote Originally Posted by mathmari View Post
    Does it hold that $P(NN\mid D)=P(N\mid D)^2$ and $P(NN\mid D^C)=P(N\mid D^C)^2$ ?
    If the two tests are independent, yes.

    Quote Quote:
    If yes, does it hold that $P(N\mid D)=0,05$ and $P(N\mid D^C)=0,95$ ?
    Yes.

    Quote Quote:
    That would mean that we get $$P(NN)=0,05^2\cdot 001+0,95^2\cdot 0,99$$ Is this correct?
    Except for the typo where you meant "0.01" and not "0.001", yes.

    You know you have not yet answered either question, right?

  4. MHB Master
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    #3 Thread Author
    Quote Originally Posted by tkhunny View Post
    Except for the typo where you meant "0.01" and not "0.001", yes.
    Ah ok!

    So, we have that $$P(NN)=0.05^2\cdot 0.01+0.95^2\cdot 0.99=0.8935$$


    Quote Originally Posted by tkhunny View Post
    You know you have not yet answered either question, right?
    For the first question we have the following:
    $$P(\text{at least one positive})=1-P(\text{no positive})=1-P(NN)=1-0.8935=0.1065$$
    Is this correct?

    At the suggested solution the result is $0, 093661972$. Can that be?


    For the second question we have the following:
    \begin{align*}P(PP)&=P(PP\cap D)+P(PP\cap D^C) \\ & =P(PP\mid D)\cdot P(D)+P(PP\mid D^C)\cdot P(D^C)\\ & =P(P\mid D)^2\cdot P(D)+P(P\mid D^C)^2\cdot P(D^C)\\ & =0.95^2\cdot 0.01+0.05^2\cdot 0.99 \\ & =0.0115\end{align*}

    Again the suggested result is different. It is $0, 784782609$.

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    #4
    Isn't the question for P(D|at least one positive) respectively P(D|both positive)?

  6. MHB Master
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    #5 Thread Author
    Quote Originally Posted by I like Serena View Post
    Isn't the question for P(D|at least one positive) respectively P(D|both positive)?
    Ah yes!

    So, we have the following: $$P(D|\text{at least one positive})=\frac{P(D\cap \text{at least one positive})}{P(\text{at least one positive})}=\frac{P( \text{at least one positive}\mid D)\cdot P(D)}{0.1065}=\frac{P( \text{at least one positive}\mid D)\cdot 0.01}{0.1065}$$ Is everything correct so far? To what is $P( \text{at least one positive}\mid D)$ equal?

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    #6
    Yep! And P(at least one positive|D)=1-P(NN|D).

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    Yep! And P(at least one positive|D)=1-P(NN|D).
    I see!! Thank you!!

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