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I've been practicing probability for a while and now I encountered a math problem which I don't know how to approach.

One integer is selected randomly from the set $[1,50]$ . What is the probability that the selected will satisfy the following inequality: $29 \leqslant 7n+3 \leqslant 99$

I don't know how to begin. Could someone help me with this and reference a link or a web page where I can learn more about math problems like this ?

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    $\begingroup$ 4<=n<=13, prob=10/50=.2 $\endgroup$ – Amruth A Jul 15 '16 at 10:12
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The inequality can be written as $$26\leq 7n\leq 96,$$ so that we require the integer $n$ to be between $4$ and $13$ inclusive. There are $10$ such integers. The probability of selecting one of those $10$ out of $50$ is simply $10/50=1/5$.

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The problem most likely intended to imply that the integer is selected uniformly randomly from $[1,50]$. If so, all $50$ outcomes are equiprobable, and you need to count the number of favourable outcomes and divide it by the total number $50$ of outcomes. What is the least integer $n$ with $29\le7n+3$? What is the greatest integer $n$ with $7n+3\le99$? Hence, how many integers satisfy both inequalities?

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