Probability of choosing two numbers so they differ by at least 2 A box has $10$ balls numbered $1,2, \dots, 10.$ A ball is picked at random and then a second ball is picked at random from the remaining nine balls. Find the probability that the numbers on the two selected balls differ by at least $2$.
I found the correct answer $4/5$ but noticed that the solution can be written as $$\frac{\binom{9}{2} }{ \binom{10}{2} }.$$
My method was not the simplest and it appears that there is a much easier way of obtaining the solution. It's supposedly the number of pairs for $9$ balls divided by the number of pairs for $10$ balls, without caring about the order. 
Could someone explain how you would arrive at this conclusion?
 A: Pick two numbers from $1,2,\ldots,9$ and increase the largest number by one. The resulting pair of numbers will always differ by at least 2 and be numbers in the range $1,\ldots,10$.
This gives a bijection between the problem of choosing two numbers of the 9 and the problem of choosing two numbers of the 10 such that they differ by at least 2. It is important that it is a bijection, i.e. two (or more) pairs of the 9 can't map to the same pair of the 10 that differ by at least two and every pair of the 10 that differ by at least 2 gets mapped to by some pair of the 9. In this case it is trivial to see that $(a,b) \mapsto (a,b+1)$ is indeed a bijection.
A: You could consider a possibility space diagram in the form of a $10*10$ square, where the leading diagonal consisting of $(1,1), (2,2)$ and so on are deleted. Then there are 90 equally likely outcomes. Of these, 18 have a difference of 1, so there are 72 remaining where the difference is 2 or more. Therefore the answer is $\frac{72}{90}=\frac 45$
A: Hint: Simply consider a $10*10$ square diagram, and you choose a square at random. The $x$ coordinate will be your first draw, and the $y$ coordinate will be your second draw. You simply have to delete the main diagonal, since you can't draw $2$ with the same number. You have $100-10=90$ squares, and in $18$ cases, you can choose $(x,y)$, so that $x+1=y$, or $x-1=y$. Since $\frac{90-18}{90} = \frac45$, that is your corrent answer.
