As has been pointed out, this probability is $1 - {{9800 \choose 200} \over {10000 \choose 200}}$. But you may be wondering how to get a numerical value for the ${9800 \choose 200} \over {10000 \choose 200}$ term.
One way is to expand those binomial coefficients in terms of factorials, giving
$$ {{9800! \over 200! 9600!} \over {10000! \over 200! 9800!}} $$
Now canceling and rearranging gives
$$ {9800! / 9600!} \over {10000! / 9800!} $$
and you can explicitly write out the products to get
$$ {9800 \times 9799 \times \cdots \times 9601 \over 10000 \times 9999 \times \cdots \times 9801} $$.
This is still problematic in that both the numerator and denominator are very large integers. But you can rewrite this as
$$ {9800 \over 10000} \times {9799 \over 9999} \times \cdots \times {9601 \over 9801} $$
and now each factor is a rational number a bit less than 1. This can be computed, for example, in Python as
x = 1
for i in range(9601, 9801):
x *= i/(i+200)
which returns $0.0168786$.
Alternatively, using Stirling's approximation $n! \approx \sqrt{2 \pi n}(n/e)^n$ we get
$$ {9800!^2 \over 9600! 10000!} \sim {(2\pi \times 9800) (9800/e)^{9800 \times 2} \over \sqrt{2\pi \times 9600} (9600/e)^{9600} \sqrt{2\pi \times 10000} (10000/e)^{10000}} $$
and simplifying a bit this is
$$ {9800 \over \sqrt{9600 \times 10000}} {9800^{19600} \over 9600^{9600} 10000^{10000}}. $$
or
$$ {9800 \over \sqrt{9600 \times 10000}} \left( {9800 \over 9600} \right)^{9600} \left({9800 \over 10000} \right)^{10000} $$
which evaluates to $0.0168786$ as well, for example in Python
import math
from math import sqrt
(9800/9600)**9600 * (9800/10000)**10000 * 9800/sqrt(9600*10000)
Finally, if those powers are too large for whatever you're doing the computation with (they're roughly $10^{86}$ and $10^{-88}$, respectively), you can rewrite as
$$ \exp \left( \log 9800 - {1 \over 2} \log 9600 - {1 \over 2} \log 10000 + 9600 \log (9800/9600) + 10000 \log(9800/10000) \right).$$
