The complement of a complement is the original event, so (A′)′=A
True — "not (not A)" is just A. If you step out of a slice and then step out of that, you land back inside the original slice.
If two events are mutually exclusive, they must be complements of each other
False — mutually exclusive only means they don't overlap (A∩B=∅). To be complements they must also be exhaustive (A∪B=S). "Roll a 1" and "roll a 2" never overlap but leave 3,4,5,6 uncovered.
P(A)+P(A′)=1 holds even when P(A)=0
True — an impossible event has complement equal to all of S, so 0+1=1. The identity is universal; it never depends on A being "likely."
For any events A and B, P(A′∩B′)=1−P(A)−P(B)
False — that only works if A and B are mutually exclusive. In general use De Morgan: A′∩B′=(A∪B)′, so P(A′∩B′)=1−P(A∪B), and P(A∪B) may include an overlap you subtracted twice. See 1.3.03-Addition-rule-for-probabilities.
If A⊆B then A′⊆B′
False — the inclusion flips. If A sits inside B, then "not B" (a smaller outside region) sits inside "not A" (a larger one), so B′⊆A′.
"At least one head" and "at least one tail" are complementary events
False — in three flips, HTH has both a head and a tail, so both events happen at once. Their complements are "all tails" and "all heads" respectively; those are the disjoint ones.
P(A′∪B)=1−P(A)+P(A∩B)
This can be true or false depending on the numbers, but the safe general form is P(A′∪B)=1−P(A∩B′) via complement of A∩B′. Never assume a shortcut formula without checking overlaps.
The complement of "rain" is "not rain," which includes cloudy, snowy, foggy — not just sunny. 1−P(rain) gives P(not rain), a bigger bucket than "sunny."
"A and A' are mutually exclusive, so I can add P(A) + P(A') into other event sums too."
The identity P(A)+P(A′)=1 only describes A with its own complement. You cannot fold A' into an unrelated sum; check mutual exclusivity fresh for any new pair.
"P(defect caught) = 0.95, so P(no defects exist) = 0.05."
The complement of "caught" is "not caught" (undetected defect), not "no defect exists." The 0.05 is the miss rate, a different statement entirely.
"P(A) = 0.6 and P(B) = 0.7, and since these are complements, they should add to 1 but 0.6 + 0.7 = 1.3."
They aren't complements — two numbers summing above 1 is your proof they can't be A and A'. Complements are forced to sum to exactly 1; anything else means unrelated events.
"For a die, P(not 6) = 1 − P(6) = 1 − 6 = −5."
The writer used the face value 6 instead of the probabilityP(6)=1/6. Always subtract a probability from 1, never a count or a face value.
"P(at least one six in two rolls) = 1 − 1/6 = 5/6."
Wrong complement value. "No six in two rolls" is (5/6)2=25/36, so the answer is 1−25/36=11/36. The complement of "at least one" is "none," and "none" must account for all trials. See 1.3.12-At-least-one-problems.
"P(A') = 1 − P(A) needs A and A' to be independent."
No — it needs them mutually exclusive and exhaustive, which they always are by definition. Independence is a separate idea about A and another event; see 1.3.09-Independent-events.
Why does P(A′)=1−P(A) require the addition rule as a step? :::Because we split P(S)=P(A∪A′) into P(A)+P(A′), and that split is only legal when the two parts don't overlap — which is exactly the addition rule's condition, met since A∩A′=∅.
Why is the complement handy for "at least one" problems? :::Because "at least one" spreads across many messy cases, but its complement "none" is a single clean case (every trial fails), so 1−P(none) replaces a long sum with one product. This is the engine of 1.4.05-Binomial-probability's P(X≥1)=1−P(X=0).
Why can't we say the complement of "rolling an even number" is "rolling a prime"? :::Because a complement must be defined as everything else in S. "Not even" means odd (1,3,5), while "prime" (2,3,5) both overlaps evens and misses 1 — it fails to partition the sample space. Complements come from S, not from another named property.
Why does the complement rule stay valid on a continuous sample space (like a spinner)? :::Because A and A′ still tile the whole space with no overlap, so their probabilities (now areas or lengths) must fill up to 1. The proof used no counting — only exhaustiveness and disjointness, which survive in the continuous world.
Why must A′ always be a subset of the same sample space S? :::Because A′=S−A is literally "the leftovers of S." An outcome outside S was never possible, so it can't sit in A′. This ties directly to 1.3.01-Sample-space-and-events.
A′=∅, the impossible event, and P(A′)=1−1=0. Nothing is left over once your event is everything.
If A=∅ (the impossible event), what is A′ and P(A′)?
A′=S, the certain event, and P(A′)=1−0=1. The complement of "never" is "always."
Can A and A′ ever both have probability 1/2?
Yes — e.g. one fair coin flip with A= heads. Nothing forbids an even split; the rule only fixes their sum at 1, not each value.
Can P(A) and P(A′) both be nonzero yet one event have zero outcomes?
In a discrete space, no — a probability comes from its outcomes, so zero outcomes forces P=0. In a continuous space a single point can have probability 0 while A′ still equals almost all of S; the outcome "exists" but carries no probability mass.
If an outcome sits on the "boundary" and you're unsure whether it belongs to A, does the complement rule break?
No — the rule doesn't care which side it lands on, only that it lands on exactly one side. Define A precisely first (put the boundary case in or out), and A′ automatically takes the rest so the sum stays 1.
What is the complement of "at least two heads in three flips"?
"At most one head," i.e. "zero or one head." The complement of "≥2" is "<2," which is "≤1" — flip the direction and include the boundary correctly.