Csdrhrt

First of all I know this question has been solved, but none of the online forums really explain it well, I am really confused with this question.

1. 
   

   Suppose you choose at random a real number X from the interval $[2; 10]$.

   
   
   

   (a) Find the density function $f(x)$ and the probability of an event $E$ for this experiment, where $E$ is a subinterval $[a; b]$ of $[2; 10]$.

   
   
   

   (b) From (a), find the probability that $X > 5$, that $5 < X < 7$, and that
   $X^2 -12X + 35 > 0$.

   
   

For instance for part a, $f(x)= 1/8$. Why? If $X$ is chosen from the interval $[2;10]$, which means $2le xle 10$, from here there are nine numbers, so why is it $1/8$?

From the first sentence we obtain that there is given a random variable $X:Omegato [2;10]$ such that for any Lebesgue measurable $Asubset[2;10]$ one has
$$mathbb{P}(Xin A)=frac{int_A1dx}{int_{[2;10]}1dx}=frac{|A|}{8},$$
where by $|A|$ I denote the one dimensional Lebesgue measure.

a) By definition a density function of a random variable $X$ is a Lebesgue measurable function defined on $X(Omega)$ such that for
$$mathbb{P}(Xin A)=int_Af(x)dx.$$

Getting back to our problem we look for $f$ such that
$$frac{|A|}{8}=int_Af(x)dx.$$
Clearly $f(x)equivfrac{1}{8}$, satisfies the above condition. To answer the second part of a) we compute
$$mathbb{P}(Xin[a;b])=int_{[a;b]}f(x)dx=frac{b-a}{8}.$$

b) We have
$$mathbb{P}(X > 5)=int_{(5;10]}f(x)dx=frac{10-5}{8}=frac{5}{8},\
mathbb{P}(5 < X < 7)=int_{(5;7)}f(x)dx=frac{7-5}{8}=frac{1}{4}.\
$$
To compute the last probability we first solve the inequality $x^2-12x+25>0$ for $xin[2;10]$. The roots of the quadratic function are $x_{1,2}=6pmsqrt{6}$, so the solution is $xin[2;6-sqrt{6})cup(6+sqrt{6};10]$. Finally
$$mathbb{P}(X^2-12X+25>0)=mathbb{P}(Xin[2;6-sqrt{6})cup(6+sqrt{6};10])=int_{[2;6-sqrt{6})cup(6+sqrt{6};10]}f(x)dx=frac{1}{8}(6-sqrt{6}-2+10-(6+sqrt{6}))=frac{1}{8}(8-2sqrt{6})=1-frac{sqrt{6}}{4}.$$

You are dealing with continuous random variable, not a discrete one.

The interval is a real number line segment of length $8$.   The uniform selection of REAL numbers from this interval will thus require a probability density of $tfrac 1 8$, in order that this integral is unity.

$$int_{[2;10]} f_X(x)operatorname d x = int_2^{10} frac 1 8 operatorname d x = 1$$

This leads into (b):

$$mathsf P(aleq Xleq b) = int_a^b tfrac 1 8operatorname d x = frac {b-a}{8} qquad mbox{iff }Big[ 2leq aleq bleq 10Big]$$

PS:

If $X$ is chosen from the interval $[2;10]$, which means $2≤x≤10$, from here there are nine numbers, so why is it $1/8$ ?

As mentioned above, you are dealing with a real number interval, so there are many, many, many more real numbers in the interval than those integers.   As to why the interval is of length $8$ when there are nine integers in the interval, refer to what is known as the "Fence Post Error".

Consider this: If I have nine fence posts placed in a line one metre apart, what length of fencing do I need?