Complement Rule vs Addition 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.

The general addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, where $\cup$ is the union sign, read "A or B", and $\cap$ is the intersection sign, read "A and B". It holds for every pair of events, with nothing to check first. The subtraction is there because any outcome living in both events was counted once inside $P(A)$ and a second time inside $P(B)$, so it has to come back out once. "Or" is inclusive: $A \cup B$ means at least one of the two, including the outcomes where both happen.

Draw one card from a standard 52-card deck. There are 13 hearts and 4 kings, and exactly one card, the king of hearts, is both. So $P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} \approx 0.3077$. Listing them agrees: 13 hearts plus the 3 kings that are not hearts is 16.

"The two events are different, so add them: $13/52 + 4/52 = 17/52$." That counts the king of hearts twice, and the list has only 16 cards in it. Plain addition is legal only when $P(A \cap B) = 0$, which is what mutually exclusive events give you: they share no outcome at all, so the subtracted term was zero anyway. The plain sum is a consequence of the general rule, never the rule itself.

For three events the pattern extends: add the three single probabilities, subtract the three pairwise intersections, then add the triple intersection back. Roll one fair die with $A = \{2, 4, 6\}$, $B = \{4, 5, 6\}$ and $C = \{2, 3, 5\}$, so the pairs hold 2, 1 and 1 outcomes and no outcome is in all three. The rule gives $\frac{3 + 3 + 3 - 2 - 1 - 1 + 0}{6} = \frac{5}{6} \approx 0.8333$, and the union is $\{2, 3, 4, 5, 6\}$, five outcomes. Reverse those signs, adding the pairs and subtracting the triple, and you get $13/6 \approx 2.17$. A probability above 1 is the tell.

Unions of events are topic 2.7.