Probability: How to get to a percentage? Exercice: We have a class of $16$ students with $2$ of them not having phones. Every class in the school is like that ($14$ with phones, $2$ without). How many students in the school do I have to pick for the probability of having at least one student without a phone to be at $99,9\%$ ?
Answer: $52$
I have no Idea how to solve this, please help. 
 A: Suppose you perform selecting $n$ students. Probability of selecting all student with a phone is $0,1\%$. Selecting one student with no phone in one class is $p= {2\over 16}={1\over 8}$. So $${n\choose n}(1-p)^n\leq 0,01\implies n \geq {\log 0,001\over \log(1-p)} = {-3\over \log 0,875}=51,7313...\implies n\geq 52$$
A: The question asks for at least one student without a phone, it could be 1 student, 2 student ,... So for solving this, we can say if the probability of having at least one student without a phone is 99.9% (or higher) then the probability of ALL having phone must be 0.1% or less. If you choose one student it has $\frac{14}{16}$ probability that he has phone. So if we choose $n$ students, then the probability of all having phone is $(\frac{14}{16})^n$
so $(\frac{14}{16})^n <0.001$ by using logarithm, we get the answer $n>51.73$ so $n\geq52$

Solving the inequality by using base 10 logarithm:
$(\frac{14}{16})^n <0.001 \Rightarrow n\times\log(\frac{14}{16}) < -3 \Rightarrow  -0.05799 \times n < -3 \Rightarrow n> 51.73$
