Classical probability definition

We select two numbers, $$a$$ and then $$b$$ from the first 100 natural numbers without replacement.
Thus the total number of ordered selections is $$100\times\ 99=9900$$

We need $$P\left(a-b\ge\ 10\right)$$

$$\therefore$$ $$\left(a\ge b+\ 10\right)$$

So for every chosen $$a$$, the value of $$b$$ must be at least 10 or less than $$a$$.

The smallest possible $$a$$ satisfying the condition is 11.

For each $$a$$, the possible values of $$b$$ are:

1,2,3,…,a−10

Number of choices for $$b$$: $$a-10$$

Favourable outcomes:

$$\Sigma_{a=11}^{a=100}\left(a-10\right)$$

Let k=a-10

$$\therefore$$ $$\Sigma_{k=1}^{k=90}\left(k\right)$$

$$sum=\dfrac{\left(90\left(90+1\right)\right)}{2}=4095$$

$$P=\dfrac{4095}{9900}$$

When we simplify for coprime, we get $$\dfrac{4095}{9900}=\dfrac{91}{220}$$

$$\therefore$$ m=91 and n=220 

$$m+n=91+220=311$$

Hence, our answer is 311.