At airport security there is a bucket with 2 red balls and 8 white balls. For each person, the security guard will select a ball with replacement. People who get a red ball will be selected for further screening. What is the probability it will take more than 20 people to get the third person selected for further screening
I think that since we need more than 20 people to get the third success then having 0 red in the first 20 or 1 red in the first 20 or 2 people in the first 20 with probability of getting a red is 2/10.
So I think that the answer is $P(Y\le2)={20 \choose 0}(2/10)^0(8/10)^{20}+{20 \choose 1}(2/10)^1(8/10)^{19}+{20 \choose 2}(2/10)^2(8/10)^{18} $
Can that be a correct solution ?
Yes, your reasoning is correct.