Simple probability of $x$ being in a segment given ratio.

Question

I am new here and I have a question. So, lets say I have a line with points $A,B,C,D$ and it looks like this. I am in precalculus and for some reason I just can't remember how to do it.

<-A-----B------C------D-->

If the ratio of these segments is as follows: $AB:BC:CD = 3:4:5$

So, what is $P(B≤x≤C)$, or the probability of a dart or $x$ landing in the $BC$ segment inside the $AD$ segment. I am trying to find the theoretical probability. How would I go about this?

Thanks for your help, -Neehar

Answer 1

To make this problem more concrete, let's say that the distance between A and B is 3 units. Then, the distance between B and C is 4 units ($\frac{4}{3} \cdot 3$), and the distance between C and D is 5 units ($\frac{5}{3} \cdot 3$). So our line now looks like this:

<---|--3---|---4---|---5----|-->

<--(A)---(B)----(C)-----(D)-->

So, we can see that the total distance between A and D is 12, and the distance between B and C is 4. So, the probability that a dart lands between B and C, given that it lands between A and D, is precisely $\frac{4}{12} = \frac{1}{3}$

Answer 2

Set $AB=a$, then the total length is $\frac{12a}{3}$. What's the probability of hitting the middle segment?

Answer 3

Choose as natural unit the one that goes with $3$, $4$ and $5$. This means that the total length is $3+4+5=12$. You are looking for the probability to hit certain 4 units of length among these 12 units.