What is the probability that a passenger will reach point a in a straight line?

J

jamesss

New member
Joined
May 9, 2021
Messages
4
The bus runs from point a to point b and back. The person got on the road somewhere between point A and B. What is the probability that he will reach A in a straight line(without passing B)? The answer has to be = 1/3, but I don't really understand why. I came to the conclusion that, for example, the path from A to B = n, the person stands on x. Obviously we are satisfied with such cases: when the bus goes from the point,where person will stand to point B(without running into the person, person came to the point a few minutes after the bus ) and when the bus goes back. I derived the probability formula depending on x, p (x) = 2 (n-x)/2n. But I can't figure out what to do next...
 
Are a and A the same? How about b and B?
Why does the answer have to be 1/3?

What would happen if we interchange the labels of A and B. Would now B have the 1/3 chance?
 
jamesss said:
The bus runs from point a to point b and back. The person got on the road somewhere between point A and B. What is the probability that he will reach A in a straight line(without passing B)? The answer has to be = 1/3, but I don't really understand why. I came to the conclusion that, for example, the path from A to B = n, the person stands on x. Obviously we are satisfied with such cases: when the bus goes from the point,where person will stand to point B(without running into the person, person came to the point a few minutes after the bus ) and when the bus goes back. I derived the probability formula depending on x, p (x) = 2 (n-x)/2n. But I can't figure out what to do next...
Click to expand...
Why do you say it must be 1/3?

You have a correct expression for the probability of reaching A before B given a position x (assuming the time and location are each uniformly distributed). The next step is to integrate.

But you will get the same answer I got by a graphical method, which is 1/2. In fact, that seems obvious, since A and B can be interchanged without changing the answer. Or am I missing something in the problem?
 
I must admit I am mixed up.

The person wants to go to A. If the bus is traveling from B to A when it arrives where the person is waiting, the person will travel directly to A. If the bus is traveling in the other direction, the person will not travel directly to A. What am I missing?
 
Yes, I beat Dr Peterson to the punch line. It must be my new picture!
 
Dr.Peterson said:
Why do you say it must be 1/3?

You have a correct expression for the probability of reaching A before B given a position x (assuming the time and location are each uniformly distributed). The next step is to integrate.

But you will get the same answer I got by a graphical method, which is 1/2. In fact, that seems obvious, since A and B can be interchanged without changing the answer. Or am I missing something in the problem?
Click to expand...
The person who told me about this task, maybe he interpreted it wrong, as I either can't see why by any chance the answer must be 1/3?
 
Jomo said:
Yes, I beat Dr Peterson to the punch line. It must be my new picture!
Click to expand...
Actually, we both edited our posts shortly after the times shown; yours started out with only one line, which I saw as I posted. The actual time at which each of us independently added the mention of symmetry is not recorded. (I'd solved the problem two ways before initially posting. But who's competing?)

On the other hand, the problem is not well stated. The actual probability is 1, if the person chooses to get on the bus only when it is going in his direction.
 
It is my way of saying that you didn't look inside my brain like you usually do with me and others. It is just a joke!
 
You must log in or register to reply here.
Top