Probability distribution

Question

10 people was randomly selected in a population. The prevalence of covid in the population is 1%. What is the probability that 1 person has covid and 9 people did not have covid?

I am not sure if this is right but I think the answer is 0.

I said P(1 person selected) = 1/10 = 0.1

But again I thought I person selected does among the 1 % will only be 0.1 not among the 10 people because we dont know the status of the 10 people.

Please help!

Answer 1

We have the conditions for a binomial trial.

B - binary output - either has COVID or doesn't

I - independent - each person independent

N - fixed number of trials = 10

S - fixed success rate of 0.01.

P(k successes) = C(n,k) (p)^k (1-p)^(n-k)

where n is sample size (10), p = probability of success = 0.01

C(n,k) is combinations of n taken k at a time.

For our problem k = 1:

C(10,1) * (0.01)^1 (0.99)^9 = 0.09135