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Problem: Given an integer $x\in\left[\text{min},\text{max}\right]$

A user comes and choose a number $\left\{ n \in \mathbb{R} : \text{min}\leq\ n \leq \text{max}\right\}$. Calculate the probability that $n > x$. I tried using following


But I am not getting correct answer. My book tells me that when $\text{min}=8156$, $\text{max}=15225$ and $x=12910$, then $P(n\gt x)=0.22474$, but this is not the answer I am getting.

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Is the distribution uniform, i.e., every number has an equal probability of getting chosen? If yes, then your answer is correct. Otherwise you need to specify the distribution. – ZeroG Jul 14 '12 at 12:43
1. What does b/w mean? 2. How do you know you are not getting the correct answer? – Gerry Myerson Jul 14 '12 at 12:46
@GerryMyerson : I would expect b/w to mean between. – ZeroG Jul 14 '12 at 12:49
@GerryMyerson According to answer book when min=8156 , max=15255 and x =12910 answer should 0.22474. But my answer is not matching with that. – Vivek Goel Jul 14 '12 at 12:50
@Vivek: If the distribution is uniform, that answer ($0.22474$) is wrong; the correct answer is about $0.33$. – joriki Jul 14 '12 at 13:06

Just to add a bit to what has already been said in the comments above.

First I don't know if you have heard about a probability density function (pdf)? If not, you can read a bit about it on this Wikipedia page. A pdf is a non-negative function defined on the real numbers (lets just go with the real numbers here) satisfying that if we take the area under the graph of $f$ from $-\infty$ to $\infty$ we get $1$. We use $f$ to calculate the probabilities that an outcome is in a certain interval. So for example the probability that a outcome is any real number is $1$ because of the area condition from above. If I want to find the probability that an outcome is in an interval $[a,b]$ I would find the area under the graph from $x=a$ to $x = b$.

The the pdf $f$ tells us something about the likelihood of finding an outcome in a certain interval. If the function is constant on an interval and zero out side, like:

$$ f(x) =\begin{cases} 0 & \text{if } x < a \\ \frac{1}{b-a} & \text{if } x \in [a,b] \\ 0 & \text{if } x > b \end{cases} $$ then we say that the pdf is a uniform density function. This just means/assumes that the only possible outcomes are in the interval $[a,b]$. (So for example $b$ would be the maximum of the possible outcomes, i.e. $b = max$ in your case).

So in your case with $a = min$ and $b = max$ our expression:

$$ \frac{max - x}{b - a} $$

is exactly the area under the graph of the pdf from x to max. Hence it gives you the probability that an outcome (the number chosen) is in the interval $[x,max]$.

But this of course assumes that the outcomes are uniformly distributed. In that case you, in you specific example, indeed get (as mentioned in the comments):

$$ \frac{15255 - 12910}{15255 - 8156} \simeq 0.3303 = 33.03 \% $$

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Does line x can be rational number don't have any effect ? – Vivek Goel Jul 15 '12 at 9:39
@VivekGoel: I am not sure that I understand your question? However, if you are asking whether saying that $x$ is rational should change the probability I would say that it doesn't (I am not sure however). But, if you assumed that $x$ should be an integer, then the only change is that the denominator is $max - min + 1$ (as mentioned in other comments and the other answer. – Thomas Jul 15 '12 at 13:40

Actually the denominator should be max-min+1 as that is the total number of integer n such that min<=n<=max.

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I think we have the real numbers here. – Thomas Jul 14 '12 at 13:50
The Op starts by saying let x be an integer which made me think min and max were integers plus the symbol n is generally used for an integer. Also the three numbers in the example are all integers. However writing n belongs to R suggests your interpretation. – Michael Chernick Jul 14 '12 at 15:06
@Thomas. It is not clear. Everything points to integers except for the use of R. But even there it is perfectly acceptable to talk about integers belonging to R. – Michael Chernick Jul 14 '12 at 15:11
I agree. It isn't obvious. My first thought had also been that $x$ should be an integer. In particular because of the $n$. I also think that it is confusing that the OP says: "Given an integer..." And the after that someone chooses a number $x$. (I by the way did not vote you down). – Thomas Jul 14 '12 at 16:13

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