Intuitively it might make sense to add the probabilities, but it doesn't actually work out that way[0]. The parent used the law of total probability to solve for the probability of hitting at least once because it's simpler:

  P(hits >= 1) = 1 - P(hits = 0).

Solving for the left side of the equation directly gets trickier:

  P(hits >= 1) = P(hits = 1) + P(hits = 2) + ... + P(hits = k)
  
  P(hits = 1) = 100 chose 1 * 0.01^1 * 0.99^99
  ...
  p(hits = n) = 100 chose n * 0.01^n *0.99^(100-n)

Taking that sum where n=100 yields the same as

  1 - P(hits = 0)
  = 1 - (100 chose 0 *0.1^0 *0.99^100)
  = 1 - 0.99^100

[0] https://en.wikipedia.org/wiki/Binomial_distribution

Indendence doesn't play into it. Simple example: flipping two coins doesn't give you a (0.5 + 0.5) = 100% chance of getting heads.

The easiest way to solve "at least once"-type questions is invariably by subtracting P(!"at least once") = P("zero hits") from 1.

To do it with addition, you'd have to list all possible outcomes completely. Those would, individually, be far lower probabilities. Example:

    P("at least one "six" when rolling two dice") 
    = P("first not six, second six") + P("first six, second not six") + P("first six, second also six") 
    = (1/6 * 5/6) + (5/6 * 1/6) + (1/6 * 1/6) 
    = 11/36 

Shorter:

    1 - P("never six") 
    = 1 - P("first not six, second not six") 
    = 1 - (5/6 * 5/6) = 36/36 - 25/36 
    = 11/36

The number of possible outcomes you have to list and add together is 2^<number of trials> - 1. For two dices, that's a manageable 2^2 - 1 = 1. For 100 rockets it's 2^100 - 1 = 1.2676506E30 (- 1).

They are independent, and that's exactly why they aren't additive. Flipping a coin two times doesn't give you a 100% chance of getting heads. To expand on the intuition a little - they may well hit the same non-target spot multiple times before they hit the target at least once, in fact that is to be expected.

Great explanations, makes perfect sense with a little more thought.

And potentially add to >100%?