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I am trying to solve this problem and I've stumbled upon something I can't overcome.

Let N be a person who is infected and T a positive test result. The probability that a person is infected is P(N) = $\frac{1}{100}$. The probability $P(T^C|N) = \frac{2}{100}$ and $P(T|N^C) = \frac{5}{100}$.

I've already calculated that the probabily someone is infected if they have one positive test result is $P(N|T) = \frac{98}{593}$ by using Bayes' law.

But how could I calculate the probability that someone is infected if they went and got 2 positive test results?

I tried by starting with the same method:

$P(N|T1,T2) = \frac{P(T1,T2|N) \cdot P(N)}{P(T1,T2)}$

Sure enough, I've already got $P(N)$, so I tried to go ahead and calculate the other two parts that I need.

For the denominator I tried this:

$P(T1,T2) = P(T1,T2|N) + P(T1,T2|N^C) = P(T|N)^2 \cdot P(N) + P(T|N^C)^2 \cdot P(N^C) = (\frac{98}{100})^2 \cdot \frac{1}{100} + (\frac{5}{100})^2 \cdot \frac{99}{100}$

I tried also thinking that it can be calculated as $P(T1,T2)=P(T1) \cdot P(T2)$ which would give me $P(T)^2$, but I feel like that's incorrect.

If I can't find $P(T1,T2)$, I feel like finding $P(T1,T2|N)$ is impossible.

How should I think in order to find these two? Is my original use of Bayes' law correct? How do I calculate the two missing pieces?

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    $\begingroup$ "which is however >1" Are you sure? $\endgroup$
    – drhab
    Oct 24, 2021 at 15:30
  • $\begingroup$ @drhab Oh shoot $\endgroup$
    – Tita
    Oct 24, 2021 at 15:44
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    $\begingroup$ @Tita If the result of the first test has no impact on the rusult of the second test, then $P(T_1\cap T_2)=P(T_1)\cdot P(T_2)=[P(T_1)]^2$. Is this the situation here? I think we can assume that. Other information are not available. $\endgroup$ Oct 24, 2021 at 19:56

1 Answer 1

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To answer my own question, my original denominator assumption is correct (I miscalculated that I have $100^3$ in the denominator). Therefore:

$P(T1,T2)=P(T1,T2|N) + P(T1,T2|N^C) = P(T|N)^2 \cdot P(N) + P(T|N^C)^2 \cdot P(N^C) = (\frac{98}{100})^2 \cdot \frac{1}{100} + (\frac{5}{100})^2 \cdot \frac{99}{100}$

The total probability is:

$P(N|T1,T2) = \frac{(\frac{98}{100})^2 \cdot \frac{1}{100}}{(\frac{98}{100})^2 \cdot \frac{1}{100} + (\frac{5}{100})^2 \cdot \frac{99}{100} }$ ~ $ 0,8 $

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