Elementary probability problem

[Elbobo]

Homework Statement

A disk is cut into 6 equal pieces and labeled 1 through 6. On it, a player spins an arrow twice. The fraction ( a / b ) is formed, where a is the number of the sector where the arrow stops after the first spin and b is the number of the sector where the arrow stops after the second spin. On every spin, each of the numbered sectors has an equal probability of being the sector on which the arrow stops.

What is the probability that the fraction ( a / b ) is greater than 1?

The Attempt at a Solution

The answer is ( 15/36 ), but I'm not really sure how to approach the problem.

(6 * 5) / 2 = 15 (total number of possible outcomes; I used this as the denominator)

As for the numerator, I couldn't find a method to find quickly the amount of fractions greater than one, and I couldn't just sit there and mess around because it was a timed exercise.


EDIT: Never mind. I guess I had to sit there and mess around, but it was a lot quicker than I thought. Is there another way to do it besides looking sector by sector?

ERR, just forget about my combination method. I figured it out.

 

[Dick]

There are 36 possible outcomes for the pair (a,b). You can throw out the cases where a=b. That leaves 36-6, right? In half of them a>b and in half of them a<b. Hence?

 

[Elbobo]

Thank you :D