Probability problem

iocal said:

Hi all,

I'll cut to the chase right away. The exercise asks me to divide a line segment into two parts by selecting a point at random. Then I must find the probability that the larger segment is at least three times the shorter. We can assume a uniform distribution. How would you would on to solve this? I know that uniform distribution means all points are equilikely but I don't know how to proceed. What is the insight here?
Here is what I have done so far. Let the length of the segment be x, then because x has a uniform distribution its PDF is x and its CDF is 1/2 * x^2 .
Let z denote the small segment after we have selected our point. Then the length of the larger segment is given by x-z. So the probability that x-z is three times larger than z is given by P(x-z>3z)= 1-F_x(4z). I cannot get a numerical result here though and the answer is supposed to be 1/2. What am I missing?

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wjm11 said:

Let's use a visual approach. Divide your line segment into 4 equal segments. The probability of choosing a point in any of the 4 segments is equal and is, therefore, .25.

Which point selections meet the 3:1 requirement? The first and last (outermost/end) segments. The middle two segments fail the 3:1 requirement.

So, two segments, with a probability of .25 each, fulfill the required conditions. .25 + .25 = .5

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