Deriving the Probability That at Least One Person is COVID-19 Positive in A Group of Random People That Meet Together for Three Days I work in the public sector. To clarify, this scenario is not being used to formulate public policy.
I am trying to calculate the probability that at least one person is COVID-19 positive in a random group of 10 individuals that we need to assemble for three days.
Based on current incidence rates for my area,the probability of at least 1 person being COVID-19 positive in a group of 10 people is 19%. This statistic was retrieved from the website https://covid19risk.biosci.gatech.edu. Criteria is group size 10 with an ascertainment bias of 3. The risk level given in percentage is stated as:
This map shows the risk level of attending an event, given the event size and location.
The risk level is the estimated chance (0-100%) that at least 1 COVID-19 positive individual will be present at an event in a county, given the size of the event.
Over the course of the three days, 8 of the 10 members of the group stay the same with 2 members changing. Total group size is consistent. The group meets for 8 hours a day after which they are free to go home. Each group member's infection status can change on a daily basis.
I'm inclined to frame the solution as follows:
$$ P(one\,person\, has\, COVID) = 1 - P( no\, one\, has\, COVID)^3 $$
with:

*

*each day as an independent event

*P(one person has COVID) =.19

*group probability remains unchanged

My main question is, after the first day,does the fact that two people out of 10 change change the group probability from .19 to some other number? My thought is no. Each day, all individuals in the group have an equal chance of infection and that the probability of having COVID-19 for any individual is independent of the status of other individuals.
For reference, here is my calculation based on the scenario:
$ Let\; P(one\,person\, has\, COVID) = .19 $ and $ P( no\, one\, has\, COVID) = .81$
Let n = 3 for the number of days which yields:
$ P( no\, one\, has\, COVID) = (.81)^3 = .5314$
This represents the probability that in a group of 10 people over three separate days, the probability that no one is COVID-19 positive is 53.14%
$P(one\,person\, has\, COVID) = 1 - .5314 = .4685$
My answer is 46.85%
 A: I disagree with your analysis.  As I see it, the critical factor is the total number of (random) people involved.
As you describe it, since $2$ of the people are replaced, there is an exposure of $(12)$ people.

Based on current incidence rates for my area,the probability of 1 person being COVID-19 positive in a group of 10 people is 19%.

Let $p = $ the probability that a specific person is COVID-19 positive.
Let $q = 1 - p$.
Then $0.81 = q^{(10)} \implies q \approx 0.979 \implies p \approx (1 - 0.979) = 0.021.$
Then, the probability of at least one person being COVID-19 positive, in a group of $12$ is
$1 - q^{12} \approx 0.223$.
The problem with your approach is that exposure on each of the $3$ days are not independent events, since you do not have $10$ different people, on each of the $3$ days.
Admittedly, I am making the assumption that if someone is not COVID-19 positive on day 1, they will also not be COVID-19 positive on day 3.

As a more graphic illustration:
which is more dangerous:

*

*taking a bus ride with the same $10$ people, $3$ days in a row:


*taking a bus ride, 3 days in a row, where there are $10$ different people on each of the $3$ days.
