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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 $5n+3\leq14$?

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closed as off-topic by Yes, Adam Hughes, Sami Ben Romdhane, drhab, martini Jul 30 '14 at 9:18

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2 Answers 2

First solve for your inequality.

$5n + 3 \leq 14$
Subtract 3 from both sides.
$5n \leq 11$
Divide both sides by $5$.
$n \leq 11/5$
$n \leq 2.2$
Since $n$ has to be a positive integer less than or equal to $10$, there are 10 possible number numbers which n could be. ($n \in \{1,2,3,4,5,6,7,8,9,10\}$).
From the inequality, we can see that $n$ can only be $1$ or $2$. Since only $2$ out of $10$ numbers can fulfill this inequality, the probability that an appropriate $n$ is chosen is $\frac{2}{10} = \frac{1}{5} = .20 = 20\%$

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Thank you very much for helping –  cary Sep 12 '12 at 21:33
This of course assumes select at random means n has a uniform distribution on the integers from 1 to n. –  Michael Chernick Sep 13 '12 at 2:38

Since $n\in\{1,2,3,4,5,6,7,8,9,10\}\implies (5n+3)\in \{8,13,18,23,28,33,38,43,48,53\}$

Now Since $5n+3$ takes only two values out of ten which are less than or equal to $14$

$\implies P(5n+3\leq 14)=\frac{2}{10}=0.2$ (Assuming each number from $1$ to $10$ is equally likely to be chosen)

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