Complement rule

Written in symbols, $P(A^c) = 1 - P(A)$, where $A^c$ is read A complement and names the event that $A$ fails to occur. It is exact for every event, needing no independence check and no equal-likelihood assumption, because the reason is structural: $A$ and $A^c$ share no outcomes and between them cover the entire sample space, so their probabilities must add to exactly 1.

Let $X$ be the number of pets in a household, with $P(0) = 0.42$, $P(1) = 0.28$, $P(2) = 0.19$, $P(3) = 0.08$, and $P(4) = 0.03$. The chance a household owns at least one pet is $1 - 0.42 = 0.58$. Adding the other four gives $0.28 + 0.19 + 0.08 + 0.03 = 0.58$ as well, four additions in place of one subtraction. Compute whichever side is cheaper to count.

The misreading shows up as soon as the sample space has more than two pieces. Suppose a flight is on time with probability 0.78, delayed with probability 0.19, and cancelled with probability 0.03. The sentence "it is 19 percent likely to be delayed, so it is 81 percent likely to be on time" is wrong. The quantity $1 - 0.19 = 0.81$ is the chance the flight is not delayed, and it counts the cancellations too. A complement is everything else in the sample space, never the one alternative you had in mind.

The rule survives conditioning, but only on the side of the bar where it belongs. If 30 of the 150 students who bike to school are late, then $P(\text{late} \mid \text{bikes}) = 0.20$ and $P(\text{not late} \mid \text{bikes}) = 0.80$, because $P(A^c \mid B) = 1 - P(A \mid B)$ holds for any fixed condition $B$. What it will not give you is $P(A \mid B^c)$, the rate among students who do not bike: 34 of those 850 are late, so it is $34/850 = 0.04$, nowhere near 0.80. Complementing the event is legal, complementing the condition is a different question. Topic 2.4 states the rule; the at least one shortcut is where it earns most of its keep.