Probability that all indices divisible by 100 in a hash table with linear probing are occupied
This problem may be reduced to the following simple probability problem with concrete numbers.

Assume you have an urn with 8 black and 12 white balls, total 20 balls. You randomly draw 15 balls from the urn one by one without replacement. What is the probability that all black balls are selected?

How does it relate to the original problem? Occupying slots one by one is equivalent to drawing balls without replacement. Both number of slots and balls decrease. "Drawing a ball" corresponds to "eventually occupying a slot" after linear probing (searching). 

We have address space of $10^6$ slots equivalent to having $10^6$ balls in an urn. Then you have positions with indices divisible by 100 equivalent to having $10000$ $(=10^6/100)$ black balls. $500000$ positions are occupied means that you draw $500000$ balls. Finally, the requirement "all positions with indices divisible by 100 $(=10^6/100)$ are occupied" means "all $10000$ black balls must be selected". 

Thus it is the hypergeometric distribution that describes the probability of $k$ successes in $n$ draws, without replacement, from a finite population of size $N$ that contains exactly $K$ successes, wherein each draw is either a success or a failure (black or white). 
The probability is equal to

$$ P(k \text{ successes})= \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

So, if we take urn-balls model then we have
$$ P(8 \text{ black balls})= \frac{\binom{8}{8}\binom{20-8}{15-8}}{\binom{20}{15}}$$

As for the original problem then we have 
$$ P(10000 \text{ occupied slots})= \frac{\binom{10000}{10000}\binom{10^6-10000}{500000-10000}}{\binom{10^6}{500000}}$$

Now, our numbers in the original problem is too large to compute the binomial coefficients. You could get around the problem using Stirling's approximation.