I'm attempting this problem from "The Probability Tutoring Book" by Carol Ash.
15% of the population is left handed. If you stop people on the street, what is the probability that
a)It takes at least 20 people to get a left handed person.
My attempt:
The probability of at least asking 20 people is going to be
$$P( \geq 20\ people) = 1 – P( \leq 19\ people) $$ Which is true because of the complement rule.
We now just need to find the $P( \leq 19\ people)$
$$P( \leq 19\ people) = P( 0\ people) + P( 1\ person) +P( 2\ people)+ … +P( 19\ people)$$
$$P( \leq 19\ people) = \sum_{i=0}^{19}{(.15)*(.85)^i}$$
So putting it all together,
$$P( \geq 20\ people) = 1 – \sum_{i=0}^{19}{(.15)*(.85)^i} \approx 0.039 $$
However, according to the solutions at the back of the book:
$$P( \geq 20\ people) = P(first\ 19\ people\ are\ right\ handed)$$
$$P( \geq 20\ people) = (.85)^{19} \approx .046$$
Wy don't our two answers concur? Am I missing something? Is the answer key wrong?
Best Answer
"It takes $20$ people to get a left-handed person" means that the $20^{\text{th}}$ person will be left-handed. You are right by applying the complement: $$P( \geq 20\ \text{people}) = 1 - P( \leq 19\ \text{people})$$
However, observe carefully here that, $ P(\text{asking} \ 19 \ \text{people})$ is actually $(.15) \times (.85)^{18}$, because asking $19$ people means that the first $18$ people will be right-handed, and the $19^{\text{th}}$ will be left-handed.
Therefore, you just have to fix this in your formula: $$P( \leq 19\ \text{people}) = \sum_{i=0}^{18}{(.15)*(.85)^i} $$
P.S: Notice also that we start summing from $i = 0$, which is the case where we meet a left-handed person on the first try (for which the probability is exactly $0.15$).
Summing up, the answer will be $$P( \geq 20\ \text{people}) = 1 - \sum_{i=0}^{18}{(.15)*(.85)^i} \approx 0.0455 $$