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

This page is the practice arena for the parent complement rule. We will hunt down every kind of situation the rule can be thrown into, then work one full example for each. If you have never seen the notation, read the parent first — but every symbol used here is re-explained the moment it appears.

Look at the amber slice: it is $A$. The cyan slice is everything else, $A^{\prime }$. The whole disc is the sample space $S$ — all possible outcomes — and its area is fixed at $1$. Growing the amber slice shrinks the cyan slice by exactly the same amount. That is the complement rule, drawn.

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The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are each tagged with the cell they cover, so together they touch all of them.

Cell	Case class	What makes it tricky	Example
C1	Simple single event	none — warm-up	Ex 1
C2	"At least one" (repeated trials)	direct count explodes	Ex 2
C3	Degenerate: $P(A)=0$	complement is certain	Ex 3
C4	Degenerate: $P(A)=1$	complement is impossible	Ex 3
C5	Complement mis-identified	"not rain $\ne$ sunny" trap	Ex 4
C6	Real-world / reliability	failure rate wording	Ex 5
C7	Limiting behaviour ($n\to \infty$)	does chance $\to 1$?	Ex 6
C8	Exam twist: complement of a complement	double negation	Ex 7
C9	Fraction bookkeeping	messy denominators	Ex 8

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Worked Examples

Forecast: guess now — is the answer bigger or smaller than $\frac{1}{2}$? Jot it down.

1. Name the event. Let $A=$ "roll a number greater than $4$", i.e. the outcomes $\{5,6\}$. Why this step? We choose $A$ to be the smaller, easier set so its complement (what we actually want) drops out by subtraction.
2. Count $P(A)$. There are $6$ equally likely faces, $2$ of them favourable, so $P(A)=\frac{2}{6}=\frac{1}{3}$. Why this step? Equally likely outcomes let us use $P=\frac{\text{favourable}}{\text{total}}$.
3. Apply the rule. The thing we want, "not greater than $4$", is exactly $A^{\prime }$: $P(A^{\prime })=1-P(A)=1-\frac{1}{3}=\frac{2}{3}.$ Why this step? $A$ and $A^{\prime }$ fill the whole pie, so subtracting from $1$ gives the leftover slice.

Verify: count directly — "not greater than $4$" is $\{1,2,3,4\}$, that's $4$ outcomes, $\frac{4}{6}=\frac{2}{3}$. ✓ Matches.

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Forecast: more or less than $0.9$?

1. Spot the complement. "At least one head" fails only when there are zero heads, i.e. all four are tails. Call the "all tails" event $A^{\prime }$, so $A=$ "at least one head". Why this step? "At least one" has many messy cases (1, 2, 3 or 4 heads); "none" has exactly one case. Always flip to the smaller side.
2. Probability of all tails. Each flip is independent (see 1.3.09-Independent-events) with $P(\text{tail})=\frac{1}{2}$, so $P(A^{\prime })={\left(\frac{1}{2}\right)}^4=\frac{1}{16}.$ Why this step? For independent events we multiply the per-flip probabilities.
3. Subtract. $P(A)=1-\frac{1}{16}=\frac{15}{16}.$

Verify: total outcomes $=2^4=16$; only $1$ of them (TTTT) has no head, so $15$ have at least one, $\frac{15}{16}$. ✓ (This is the engine behind 1.3.12-At-least-one-problems.)

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Forecast: what feels like the answer to each — $0$ or $1$?

1. (a) Identify the impossible event. Let $A=$ "roll a $7$". A die has no face $7$, so $A$ can never happen: $P(A)=0$. Why this step? An event containing zero outcomes has probability $0$ — this is the degenerate edge case.
2. Complement of the impossible. $P(A^{\prime })=1-0=1.$ "Not rolling a $7$" is certain, which matches reality. This is cell C3.
3. (b) Identify the certain event. Let $B=$ "roll a number from $1$ to $6$". Every face satisfies this, so $P(B)=1$.
4. Complement of the certain. $P(B^{\prime })=1-1=0.$ "Not rolling a $1$–$6$" is impossible. This is cell C4.

Verify: the complement rule must map $0\leftrightarrow 1$. It does: $P(A)=0\Rightarrow P(A^{\prime })=1$ and $P(B)=1\Rightarrow P(B^{\prime })=0$. The two degenerate ends behave exactly as common sense demands. ✓

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Forecast: is the student right, partly right, or wrong?

1. State what the rule actually gives. The complement of "rain" is "not rain", not "sunny": $P({\text{rain}}^{\prime })=1-0.3=\mathrm{0.7.}$ Why this step? $A^{\prime }$ is everything in $S$ outside $A$ — here that bundle is cloudy $+$ snowy $+$ foggy $+$ sunny $+\dots$, all lumped together.
2. Why "sunny" is too small. "Sunny" is only one slice of the non-rain region. So $P(\text{sunny})\le P(\text{not rain})=0.7,$ and it is $0.7$ only if sunny is the only alternative to rain — which is rarely stated. Why this step? The complement rule is exact; jumping from "not rain" to "sunny" silently throws away all the other weather.

Verify (a made-up split): suppose non-rain weather is sunny $0.5$, cloudy $0.15$, foggy $0.05$. Their sum is $0.5+0.15+0.05=0.7=P({\text{rain}}^{\prime })$. ✓ The $0.7$ is the total, not the sunny part.

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Forecast: tiny (near $0$) or moderate?

1. Get the per-part slip probability via complement. Let $C=$ "sensor catches the fault", $P(C)=0.92$. Then $P(C^{\prime })=1-0.92=\mathrm{0.08.}$ Why this step? The problem gives the catch rate; the slip rate is its complement.
2. Both slip (independent). $P(\text{both slip})=P(C^{\prime })\times P(C^{\prime })=0.08\times 0.08=\mathrm{0.0064.}$ Why this step? Independent failures multiply — the second slip does not depend on the first.

Verify: $0.08^2=0.0064=0.64\%$, a small but non-zero risk — plausible for a duplicated defect. Units check: a probability, dimensionless, and $0<0.0064<1$. ✓

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Forecast: does the chance settle near $\frac{1}{2}$, or climb toward $1$?

1. General formula. As in Ex 2, the only failure is all tails: $P(\text{at least one head})=1-{\left(\frac{1}{2}\right)}^n.$ Why this step? One clean complement handles any $n$ at once.
2. Watch the tail. As $n\to \infty$, ${\left(\frac{1}{2}\right)}^n\to 0$, so $P(\text{at least one head})\to 1-0=1.$ Why this step? Halving repeatedly drives the "all tails" slice toward nothing, so the amber slice swells to fill the pie.
3. Value at $n=10$. $1-{\left(\frac{1}{2}\right)}^{10}=1-\frac{1}{1024}=\frac{1023}{1024}\approx \mathrm{0.99902.}$

The plot shows the amber "all tails" bar shrinking toward $0$ while the cyan "at least one head" bar climbs toward $1$. The complement rule ties them together at every $n$.

Verify: $\frac{1023}{1024}=0.999023\dots$, just below $1$, consistent with the limit being $1$ but never quite reached for finite $n$. ✓

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Forecast: should this collapse back to something familiar?

1. Apply the rule once. $P(A^{\prime })=1-0.35=\mathrm{0.65.}$ Why this step? First negation: "not $A$".
2. Apply the rule again. "$(A^{\prime }{)}^{\prime }$" means "not (not $A$)", which is just $A$ back again: $P((A^{\prime }{)}^{\prime })=1-P(A^{\prime })=1-0.65=\mathrm{0.35.}$ Why this step? Two negations undo each other — geometrically, flip the pie inside-out twice and you are back to the original slice.

Verify: $P((A^{\prime }{)}^{\prime })=P(A)=0.35$. ✓ Double-complement is the identity, exactly like turning a light off then on.

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Forecast: more or less than $\frac{1}{2}$?

1. Choose the easy side. Let $A=$ "green", with $4$ favourable of $12$: $P(A)=\frac{4}{12}=\frac{1}{3}.$ Why this step? Counting one colour ($4$) beats counting the other two ($5+3=8$)... barely, but it also shows the fraction reduces cleanly.
2. Complement. $P(A^{\prime })=1-\frac{1}{3}=\frac{2}{3}.$

Verify: direct count of "not green" $=5+3=8$ out of $12=\frac{8}{12}=\frac{2}{3}$. ✓ Denominators reconcile.

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Recall Rapid self-test (cover the answers)

Every scenario cell has one quick check. Reveal after guessing.

Connections

• Complementary events — P(A') = 1 − P(A) — the parent rule these examples drill.
• 1.3.01-Sample-space-and-events — every $A^{\prime }$ lives inside the sample space $S$.
• 1.3.03-Addition-rule-for-probabilities — the identity $P(A)+P(A^{\prime })=1$ comes from it.
• 1.3.09-Independent-events — used to multiply per-trial slips in Ex 2, 5, 6.
• 1.3.12-At-least-one-problems — Ex 2 & 6 are its core pattern.
• 1.4.05-Binomial-probability — $P(X\ge 1)=1-P(X=0)$ generalises Ex 6.