Vtyjkyuk

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
$$mathbbP(Xin A)=fracint_A1dxint_[2;10]1dx=frac8,$$
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
$$mathbbP(Xin A)=int_Af(x)dx.$$

Getting back to our problem we look for $f$ such that
$$frac8=int_Af(x)dx.$$
Clearly $f(x)equivfrac18$, satisfies the above condition. To answer the second part of a) we compute
$$mathbbP(Xin[a;b])=int_[a;b]f(x)dx=fracb-a8.$$

b) We have
$$mathbbP(X > 5)=int_(5;10]f(x)dx=frac10-58=frac58,\
mathbbP(5 < X < 7)=int_(5;7)f(x)dx=frac7-58=frac14.\
$$
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 $x_1,2=6pmsqrt6$, so the solution is $xin[2;6-sqrt6)cup(6+sqrt6;10]$. Finally
$$mathbbP(X^2-12X+25>0)=mathbbP(Xin[2;6-sqrt6)cup(6+sqrt6;10])=int_[2;6-sqrt6)cup(6+sqrt6;10]f(x)dx=frac18(6-sqrt6-2+10-(6+sqrt6))=frac18(8-2sqrt6)=1-fracsqrt64.$$

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):

$$mathsf P(aleq Xleq b) = int_a^b tfrac 1 8operatorname d x = frac b-a8 qquad mboxiff Big[ 2leq aleq bleq 10Big]$$

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?