# Linear system with probabilities

I need help to solve this problem

an urn contains the numbers 1,2,3, ......., 2019. we draw at random, without replacement, four numbers in the order, in the urn, which we will designate by a, b, c , d. what is the probability that the system (S)

(S): ax+by=ab and cx+dy=cd

has a strictly internal solution (that is, not on the axes) in the first quadrant?

• What is meant by "the system S" and "internal solution"? – Michael Jun 10 '19 at 22:14
• I corrected internal means the first quadrant except the axes (ie x>0 and y>0) – cerise Jun 10 '19 at 22:19
• $(x,y) = \displaystyle \left(\frac{bd(c-a)}{bc-ad}, \frac{ac(b-d)}{bc-ad}\right)$ – Saketh Malyala Jun 10 '19 at 22:46
• what does that tell you about the values of a,b,c,d, that can give a nonzero solution for both x and y – Saketh Malyala Jun 10 '19 at 22:46
• one more thing is that c-a and b-d have to be the same sign as bc-ad – Saketh Malyala Jun 10 '19 at 22:55

It can easily be found that $$(x,y)=(\frac{bd(c-a)}{bc-ad},\frac{ac(b-d)}{bc-ad})$$.
As it has a 'strictly internal solution' $$c-a$$, $$b-d$$ and $$bc-ad$$ must have the same sign.
If $$c>a$$, then $$b>d$$. Under this condition, $$bc-ad$$ is always true. Similarly, if $$c and $$b, then $$bc-ad<0$$ is always true.
Now, the total number of possible relative positions of a,b,c and d under the given condition is $$2\cdot\frac{4!}{(2!)^{2}}$$ (Under each condition, the relative position of a,c and b,d are fixed). So, the total probability that the solution is strictly internal to the first quadrant is $$\frac{2\cdot\frac{4!}{(2!)^{2}}}{4!}=\frac{1}{2}$$.