Probability  what is the probability that a given tire length will lie in a lenght of interest
Walking across a street and lookign at tires I've had a thought. If a car is rolling on a road starting at point O, at each time some area of a tire is touching the road, what if after some random distance L there are 2 points  A and B, the distance from A to B is 100cm.
A tire of that car has a length of 1000cm, of that 1000cm, 10cm is painted red  R.
What would be the probablity that all of the painted area R would be inside points A and B. The area R would need to lie fully inside A & B, it can't start in outside of AB and end in AB or start in AB and end outside of AB  only in between AB.
How the probability expression would change if instead of random distance L we would assign a distance from point O to A. My first thought is that it would not change as there is another random factor  what is the starting position of a tire ( where the area R is from the starting point O ).
I've added an image drawing this question for easier visualization.

$2.90? Surely that was a typo

I thought this is an easy question for someone who is working with probability and wouldn't take more than 5 minutes. I've increased my bounty.
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Thank you for your answer, shouldn't in the very top instead of R > T be L > T?

So for example, I have a tire T with circumference of 1000, painted part R is 10, the length of a road L is 100. My intuition would suggest that the probability of R being inside of L is very low because it is a small part of the tire, in othercase if L >= T+R the probability is 1. If I calculate the probability using your formula I get that the probability is 90%, which is not what I expect, unless I understood it wrong.

Oh! I think I confused the problem a bit. I thought it was a one meter tire on a onekilometer road, but you're talking about a 10 meter tire (monster truck!) running on a one meter road with a 10 cm painted part. This answer was taken assuming the road was longer than the tire length, but rereading your question I now see that is not the case. I will rewrite the answer tomorrow.