# Distance probabilty question

1. Jun 22, 2011

### chrisphd

Consider a line segment of length 60cm, divided in half (30 cm/half). A point is randomly chosen from the first half of the line segment, and another point is randomly chosen from the second half of the line segment. What is the probability that the distance between the two points is less than 20cm?

2. Jun 22, 2011

### chiro

From the information you have said, both points are chosen from independent parts of the string.

So consider X,Y ~ Uniform(30).

Lets say P(X = x) and P(Y = y) refers to the probability of getting a value x (or y) cm's from the center of the string. You can do this because of the nature of a uniform distribution.

That means that you have to find P(X + Y < 20).

Show us some working out and I will give you more hints.

3. Jun 22, 2011

### gsal

Well...I am no physicist...I am just an engineer...so, for me this is good enough:

as an approximation...let's assume that x and y can only assume discrete values from 0 to 30...that's 31 possible values each, for a total of 31x31=961 possibilities

but when you add x and y, only 210 combinations add to < 20...

so Prob = 210/961 = 0.2185

4. Jun 22, 2011

### chrisphd

let x be distance along 1st segment, and y be distance along 2nd segment. x and y are both random and independent, so we can treat them as a pair (x,y).
How do we work out the possible values of (x,y)? Draw a graph with x axis and y axis, up to x=30 and y=30. All coordinates in this range correspond to possible (x,y) values.
Draw a straight line from (0,20) to (20,0). Everywhere along this line (eg. (20,0),(19,1) etc.) all have distance from each other equal to 20. Everywhere under this line is less than 20, and above greater than 20.

Area under this line = 20*20/2. Total area of possibility space = 30*30. Prob of distance less than 20 = 20*20/(2*30*30) = 2/9 = 0.222. This is consistent with engineers approximation so I think my method is right.

Thanks!