I’ve been doing this question but I’m a little stuck on the second part. The first part is as follows:

The probability of a randomly selected person in a population having a particular genetic trait is 0.00001. A test for this trait successfully detects it, if present, 99.9% of the time, and only returns a false positive 0.1% of the time. A person tests positive for the trait. Find the probability that they actually have the genetic trait. Give your answer to 3 significant figures.

To approach this problem, I defined the following events: A – the person has the genetic trait and B – the test detects the trait.

Based on the information in the question, we have: $ $ P(A)=0.00001$ $ $ $ P(B|A)=0.999$ $ $ $ P(B|!A)=0.001$ $

The question asks for $ P(A|B)$ and so I thought to apply Bayes: $ $ P(A|B)=\frac{P(B|A)P(A)}{P(B)}$ $

To find $ P(B)$ , I used the total probability: $ $ P(B)=P(B|A)P(A)+P(B|!A)P(!A)$ $

Substituting in the values, I got $ P(A|B)=0.00989$ to 3 significant figures. The second part goes like this:

In order to improve accuracy, individuals are instructed to take the test twice, regardless of the result of the first test. What is the probability that an individual receives a positive result from both tests? Give your answer to 3 significant figures. Find the probability, given that they have tested positive twice, that they actually have the genetic trait. Give your answer to 3 significant figures.

I’m not quite sure how to formulate this problem in terms of the events I defined previously.