A problem is as follows: 15% of the population are left-handed, if people are stopped randomly on the street, what is the probability that it takes at least 20 tries to get 3 lefties?
The solution that the author has proposed is:
$P(2 \ lefties \ in \ 19 \ tries) + P(1 \ lefty \ in \ 19 \ tries) + P(no \ lefty \ in \ 19 \ tries)$
The solution that I think would be correct is:
$P(2 \ lefties \ in \ 19 \ tries \cup 1 \ lefty \ in \ 18 \ tries \cup no \ lefty \ in \ 17 \ tries) = $
$P(2 \ lefties \ in \ 19 \ tries) + P(1 \ lefty \ in \ 18 \ tries) + P(no \ lefty \ in \ 17 \ tries) - P(1 \ lefty \ in \ 18 \ tries \ and \ 19th \ is \ a \ lefty) - P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ is \ a \ lefty) - P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ and \ 19th \ are \ lefties) + P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ and \ 19th \ are \ lefties)$
If the author is right, why? For exaxmple, how could 1 lefty in 19 tries mean "at least 20 tries to get 3 lefties"? Let's say the 19th one is a lefty, then to get 3 lefties, the most compact way would be "19th, 20th and 21st are lefties", which is not "at least 20 tries", it is "at least 21 tries". Also there seems to be intersections in that solution.
(UPDATE: The book is "The Probability Tutoring Book by Carol Ash" in case it matters.)
At least 20 means in particular at least 21. The question is not asking the probability that it takes exactly 20 tries.
Think of it is way: saying that it takes at least 20 tries is exactly like saying that out of the first 19 tries, there are strictly less than 3 lefties. This leaves the disjoint possibilities 0, 1, or 2 lefties out of 20. Do you know how to compute the final probability?
EDIT. Your solution is technically correct, but the issue is that writing the event $E =$ "it takes at least 20 tries" as the union of $A =$ "2 in 19", $B =$ "1 in 18", and $C=$ "0 in 17" makes it hard to compute. This is because these three events are not disjoint, and thus as you say, there are intersections to take into account. However, it is also true that $E$ is the union of the events $A$, $B' =$ "1 in 19", and $C'=$ "0 in 19". Now, these events are disjoint, so $$ P(E) = P(A \cup B' \cup C') = P(A) + P(B') + P(C'). $$ The added advantage is that the probability of $A, B', C'$ is easy to compute (how?).