If a positive integer $n$ is picked at random from the positive integers less than or equal to $10$, what is the probability that $5n+3\le14$?
(A) $\displaystyle\large 0$
(B) $\displaystyle\large\frac{1}{10}$
(C) $\displaystyle\large\frac{1}{5}$
(D) $\displaystyle\large\frac{3}{10}$
(E) $\displaystyle\large\frac{2}{5}$
The first thing thought of was to solve the inequality. This gave me:
$\displaystyle\large n\le\frac{11}{5}$ Then I did trial and error: I started with $n=1$ and that came out false (because 1 is not $\le\frac{11}{15}$), so I'm assuming that there is $0$ probability for n from $2\to10$ satisfying the expression. And that therefore there is $0$ probability for $5n+3\le14$ for $n$ being $0\ge n\le10$. So the answer is (A). Am I right?
[Note: I didn't start with $n=0$ because I was confused about "the positive integers less than or equal to $10\,$" and that there was controversy about 0 being neither positive nor negative. Do school assignments take this into account? Should I or should I not count $0$ as positive? - end note].