Complementary events — P(A') = 1 − P(A)

What is a Complementary Event?

Key property: A and A' are:

1. Mutually exclusive: $A \cap A' = \emptyset$ (they share no outcomes)
2. Exhaustive: $A \cup A' = S$ (together they cover the entire sample space)

Derivation: Why P(A') = 1 − P(A)

Let's build this from scratch using the axioms of probability.

Step 1: Partition the sample space Since A and A' are exhaustive: $A \cup A' = S$

Step 2: Apply the addition rule Because A and A' are mutually exclusive ($A \cap A' = \emptyset$): $P(A \cup A') = P(A) + P(A')$

Why? The addition rule says: if two events can't happen together, the probability of "either" is the sum of their individual probabilities.

Step 3: Use the total probability axiom The probability of the entire sample space is 1: $P(S) = 1$

Step 4: Combine From Step 1: $A \cup A' = S$, so $P(A \cup A') = P(S) = 1$

From Step 2: $P(A \cup A') = P(A) + P(A')$

Therefore: $P(A) + P(A') = 1$

This is always true for any event A in any probability space.

When to Use the Complement

Worked Examples

Direct approach (longer):

• Favorable outcomes: {1, 2, 3, 4, 5} → 5 outcomes
• P(not 6) = 5/6

Complement approach (faster):

• Let A = "rolling a 6"
• P(A) = 1/6
• P(A') = 1 − 1/6 = 5/6 ✓

Why this step? We already know P(6) is simple; subtracting from 1 is one calculation instead of counting five outcomes.

Direct approach (tedious):

• List all outcomes with≥1 head: HH, HHT, HTH, HTT, THH, THT, TTH → 7 outcomes
• Sample space: 2³ = 8
• P(at least 1 H) = 7/8

Complement approach (elegant):

• Let A = "at least one head"
• A' = "no heads" = "all tails" = {TT}
• P(A') = 1/8
• P(A) = 1 − 1/8 = 7/8 ✓

Why this works? "At least one" has 7 cases; "none" has only 1 case. Complement shrinks the problem.

Setup:

• Let A = "drawing a spade"
• P(A) = 13/52 = 1/4
• P(A') = 1 − 1/4 = 3/4

Why this step? We know spades = 13 cards immediately. Counting hearts + diamonds + clubs = 39 cards is more work.

Solution:

• Let A = "defect is caught"
• P(A) = 0.95
• P(A') = 1 − 0.95 = 0.05 (5% slip through)

Why this matters? In reliability engineering, knowing the failure rate (complement) is often more critical than the success rate.

Common Mistakes

Why it feels right: We think "not rain = sunny"

The fix: Not rain could be cloudy, snowy, fogy, etc. The complement is "ANY weather that isn't rain." Only use the complement rule when A' is precisely defined as "everything except A."

P(rain') = P(not rain) = 0.7 includes all non-rain weather, not just sunny.

Why it feels right: It's true that P(A) + P(A') = 1, so we want to use this sum.

The fix: A and A' are mutually exclusive and exhaustive by definition. You can't meaningfully combine them with other events using this logic. The equation P(A) + P(A') = 1 is an identity, not a calculation tool for other events.

If you have events B and C, you can't say P(B ∪ C) = P(B) + P(C) unless you verify they're mutually exclusive first.

Why it happens: We forget that subtraction is often faster than re-counting.

The fix: Always ask: "Do I already know P(A)? Is the complement easier?" If yes, use 1 − P(A).

You could count: 7 blue marbles out of 10 → 7/10.

But here's the shortcut: If the chance of red is 3/10, then the chance of "not red" is whatever is left over from 100%. So: 1 − 3/10 = 7/10.

Why does this work? Because every marble is either red or not red. There's no third option. So red + not-red = all the marbles = 100% = 1.

This trick is powerful: If you're trying to find "at least one" of something (like "at least one head in 3 coin flips"), it's way easier to calculate "zero" (all tails) and subtract from 1. Instead of counting 7 cases, you count 1 case!

Or: "Together, they make ONE"

• A and A' together = the whole sample space = probability 1

Connections

• 1.3.01-Sample-space-and-events — A' is a subset of the sample space S
• 1.3.03-Addition-rule-for-probabilities — Used to prove P(A) + P(A') = 1
• 1.3.09-Independent-events — P(A' ∩ B) = P(A') · P(B) if independent
• 1.3.12-At-least-one-problems — The killer application of complements
• 1.4.05-Binomial-probability — P(X ≥ 1) = 1 − P(X = 0) uses complements

#flashcards/maths

What is the complement of event A? :: The set of all outcomes in the sample space that are NOT in A, written A' or A^c. It represents "A does not happen."

State the complement rule formula :: P(A') = 1 − P(A), which follows from A and A' being mutually exclusive and exhaustive.

Why are A and A' mutually exclusive? :: Because they share no outcomes: A ∩ A' = ∅. An outcome cannot be both "in A" and "not in A" simultaneously.

True or False: P(A) + P(A') always equals 1 :: True, for any event A in a probability space, because A and A' partition the sample space.

A test detects a disease98% of the time. P(disease goes undetected) = ? :::1 − 0.98 = 0.02 or 2%

Concept Map

Hinglish (regional understanding)