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Let $I = [0,\ n],\ A = [a,\ b] : A \subseteq I$

Also, $0 \leq a \leq b \leq n$ and $a, b, n \in \mathbb{N}$.

I'll choose a real value $x$ uniformly out of $I$. What is the probability of the interval $B = [x, x + L]$ be completely inside the interval $A$, i.e $B \subseteq A$? $L$ is a given constant.

$x \in \mathbb{R}_{\ge 0},\ 0 \leq x \leq \ n$,

$\ L \in \mathbb{N},\ 0 < L \leq N$.

M. M
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    Think about it geometrically. What is the maximal value x can have such that still fits still into $A$? Now use the property that you have to choose x uniformly. – Quasar Nov 08 '17 at 20:30
  • By the way: What do you mean with uniformly? Out of what interval? – Quasar Nov 08 '17 at 20:30
  • It can't be over all $\mathbb{R}_{\ge 0}$: See this post: https://math.stackexchange.com/questions/14777/why-isnt-there-a-uniform-probability-distribution-over-the-positive-real-number#14779 – Quasar Nov 08 '17 at 20:32
  • What is the distribution on $(x,L)$? – copper.hat Nov 08 '17 at 20:37
  • You need to define the distribution of $L$, it has no sense if it is uniformly distributed on $N$ – Djura Marinkov Nov 08 '17 at 20:50
  • @Quasar Hey, sorry for that! I updated the question answering your comment. By uniformly I mean that every value of $I$ have an equal probability of being chosen. Is that an enough answer? My math is not good. – M. M Nov 08 '17 at 20:51
  • @DjuraMarinkov hey, L is a given constant, a natural number less or equal than $N$. Is that enough? – M. M Nov 08 '17 at 20:56
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    yes it is, $p=\frac{b-a-L}{n}$ if $L<b-a$ otherwise is 0 – Djura Marinkov Nov 08 '17 at 21:01
  • @DjuraMarinkov thanks! That makes a lot of sense, can you explain your intuition on that result please? – M. M Nov 08 '17 at 21:17

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