Probability - equation involving prime number

1. May 14, 2013

Saitama

1. The problem statement, all variables and given/known data
Consider the equation as given $\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{1}{p}$ where $x, y, z, p \in I^+$ and $p$ is a prime number & $(x,y)$ represents the solution pair then
A)probability x<y is 1/3
B)probability that x>y is 5/6
C)probability that x≠y is 2/3
D)probability that x=y is 1/6

(There can be more than one answers correct)

2. Relevant equations

3. The attempt at a solution
I am a dumb at these probability questions so I need a few hints to start with. The only thing I can think of is start with plugging a few numbers. For p=3, x=y=6, for p=5, x=y=10 but this is definitely not the way to solve the problem. I believe that there is a much better and an elegant way to solve this problem.

Any help is appreciated. Thanks!

2. May 14, 2013

Saitama

Anyone?

3. May 14, 2013

haruspex

Factor x and y into ab, ac, where b and c are defined to be coprime. What can you deduce about b and c?

4. May 15, 2013

Saitama

$(x-p)(y-p)=p^2$?

5. May 15, 2013

Curious3141

The way I did this was to rearrange the equation thus:

$$p(x+y)=xy$$

which leads to the conclusion that at least one of x or y is a multiple of p.

Consider two cases: where x is a multiple of p and y is a multiple of p.

You should be able to find the complete solution set (x,y) with this.

You will be left with 3 infinite disjoint sets of ordered pairs (x,y) of equal cardinality (they are ennumerable with an index running through all primes).

From this, you should be able to deduce which of the relationships is/are true (hint: exactly two).

6. May 15, 2013

haruspex

I was defining a to be the HCF of x and y, thus x = ab, y = ac, where b and c are coprime.
p(x+y)=xy gives p(b+c) = abc. Suppose q is a factor of b. Can it be a factor of b+c?