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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].

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Shouldn't it be n <= 11/5> –  segiddins Oct 10 '12 at 23:06
    
Right, sorry about that –  David Oct 10 '12 at 23:07
    
For your note: a number $x$ is positive if $x>0$ so $0$ is not positive. Also, you've already been corrected that you should have $n\le11/5$, so you should edit your post accordingly. I can't because deleting a single character isn't allowed, but maybe you can. –  Rick Decker Oct 10 '12 at 23:38

1 Answer 1

up vote 5 down vote accepted

Obviously you have 10 choices: $\mathbf{U}=\{1,2,\ldots, 10\}$. Two of these: $u=1$ and $u=2$ yield the required inequality, since by simple induction it does not hold for $u \geq 3$. Assuming the integer is selected uniformly at random, the correct answer is C. 0 is not a positive integer, as stated in the problem.

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I still don't understand. Why $\frac{1}{5}$? –  David Oct 10 '12 at 23:13
    
You have overall 10 outcomes, of them 2 are positive (1 and 2). Since choices are sampled uniformly at random, the probability of getting one of these two in 1 experiment is $\frac{2}{10}=\frac{1}{5}$ –  Alex Oct 10 '12 at 23:16

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