At least one probability

At least one and none are complements: every result has either zero occurrences or one or more, with nothing between them and no overlap. So $P(\text{at least one}) = 1 - P(\text{none})$, and the right-hand side stays a single term however many trials there are, while the direct route sums exactly one, exactly two, exactly three, and onward.

Roll a fair six-sided die four times. One roll misses a 6 with probability $5/6$. If the rolls are independent and each carries the same probability, four misses in a row is $(5/6)^4 = 625/1296 \approx 0.4823$, so $P(\text{at least one } 6) = 1 - 0.4823 = 0.5177$. Enumerating all $6^4 = 1296$ ordered results confirms it: 671 of them contain a 6, and $671/1296 \approx 0.5177$.

Independence and a constant success probability are conditions, not decoration, and the power step is what fails without them. Deal 4 cards from a standard deck and the chance of at least one ace is $1 - \binom{48}{4} / \binom{52}{4} = 1 - 194580/270725 \approx 0.2813$. Treating the draws as independent gives $1 - (48/52)^4 \approx 0.2740$, wrong because each card dealt changes the deck for the next.

Two wrong sentences turn up constantly. At least one is the opposite of exactly one, so subtract $P(X = 1)$: that gives $1 - 500/1296 \approx 0.6142$ for the dice above against the true 0.5177, because it throws away an outcome you wanted and keeps the one you did not. And four rolls at $1/6$ each is $4/6 \approx 0.667$: adding probabilities across trials fails its own sanity check, since six rolls would then reach 1, a certainty, while the true value is $1 - (5/6)^6 \approx 0.6651$.