# Probability that a person is infected after getting 2 positive tests.

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?

• "which is however >1" Are you sure? Oct 24, 2021 at 15:30
• @drhab Oh shoot
– Tita
Oct 24, 2021 at 15:44
• @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. Oct 24, 2021 at 19:56

To answer my own question, my original denominator assumption is correct (I miscalculated that I have $$100^3$$ in the denominator). Therefore:
$$P(T1,T2)=P(T1,T2|N) + P(T1,T2|N^C) = P(T|N)^2 \cdot P(N) + P(T|N^C)^2 \cdot P(N^C) = (\frac{98}{100})^2 \cdot \frac{1}{100} + (\frac{5}{100})^2 \cdot \frac{99}{100}$$
$$P(N|T1,T2) = \frac{(\frac{98}{100})^2 \cdot \frac{1}{100}}{(\frac{98}{100})^2 \cdot \frac{1}{100} + (\frac{5}{100})^2 \cdot \frac{99}{100} }$$ ~ $$0,8$$