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

Before we start, one picture to keep in your head the entire time — the sample space is a full bar of length $1$ , split into the slice $A$ and the leftover slice $A^{'}$ .

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

The whole bar = $1$ . Whatever height (probability) $A$ eats, $A^{'}$ gets the rest. That "rest" is the complement rule.

Level 1 — Recognition

Goal: identify $P (A)$ , then just subtract from 1.

The next picture shows the three L1 problems as bars: the coloured slice is the event $A$ we can read off instantly; the leftover (its complement $A^{'}$ ) is the answer we want.

Recall Solution 1.1

Let $A =$ "roll a $2$ ". A fair die has $6$ equally likely faces, so $P (A) = \frac{1}{6} .$ "Not rolling a $2$ " is exactly the complement $A^{'}$ . By the complement rule: $P (A^{'}) = 1 - P (A) = 1 - \frac{1}{6} = \frac{5}{6} .$ Answer: $\frac{5}{6}$ .

Recall Solution 1.2

Let $A =$ "draw a king". There are $4$ kings, so $P (A) = \frac{4}{52} = \frac{1}{13} .$ $P (not a king) = 1 - \frac{1}{13} = \frac{12}{13} .$ Answer: $\frac{12}{13}$ .

Recall Solution 1.3

"Not rain" is the complement of "rain": $P (no rain) = 1 - 0.30 = 0.70.$ Answer: $0.70$ . (Careful — this is "no rain of any kind", not "sunny". See the mistake below.)

Level 2 — Application

Goal: choose the complement because it is the easier count.

For "at least one" problems the smartest picture is a tree: each branch is a step, and the ONE branch where nothing happens is the complement. Below is Problem 2.1's coin tree — notice how only the all-tails path ( $T T T$ ) belongs to $A^{'}$ .

Recall Solution 2.1

The direct event "at least one head" has many cases. Its complement is a single case: $A = "at least one head", A^{'} = "no heads" = "all tails" = T T T .$ Each toss is independent with $P (tail) = \frac{1}{2}$ (see 1.3.09-Independent-events), so $P (A^{'}) = (\frac{1}{2})^{3} = \frac{1}{8} .$ $P (A) = 1 - \frac{1}{8} = \frac{7}{8} .$ Answer: $\frac{7}{8}$ . This is the flagship "at least one" trick.

Recall Solution 2.2

Complement $A^{'} =$ "no red at all" $=$ "both blue". First draw blue: $\frac{7}{10}$ . After removing one blue, $6$ blue remain of $9$ total, so second draw blue: $\frac{6}{9}$ . $P (A^{'}) = \frac{7}{10} \cdot \frac{6}{9} = \frac{42}{90} = \frac{7}{15} .$ $P (at least one red) = 1 - \frac{7}{15} = \frac{8}{15} .$ Answer: $\frac{8}{15}$ .

Recall Solution 2.3

"Detected" has probability $0.95$ , so "missed" (one item) has probability $1 - 0.95 = 0.05$ . Let $A =$ "at least one of the three is missed". Then $A^{'} =$ "all three are detected": $P (A^{'}) = (0.95)^{3} = 0.857375.$ $P (A) = 1 - 0.857375 = 0.142625.$ Answer: $\approx 0.1426$ (about $14.3%$ ).

Level 3 — Analysis

Goal: dissect a compound event; decide which piece to complement.

Two visual aids here. First, Problem 3.1's dice grid — every cell is one of the $36$ outcomes, and only the tiny magenta corner (sum $> 10$ ) is the complement we count. Second, a Venn reminder of the two set operations used in Problem 3.3.

Recall Solution 3.1

There are $6 \times 6 = 36$ equally likely outcomes. "At most $10$ " means sum $\leq 10$ — that's a lot of cases. Its complement is small: sum $> 10$ , i.e. sum $= 11$ or $12$ .

Sum $11$ : $(5, 6), (6, 5)$ → $2$ outcomes.
Sum $12$ : $(6, 6)$ → $1$ outcome. $P (sum > 10) = \frac{3}{36} = \frac{1}{12} .$ $P (sum \leq 10) = 1 - \frac{1}{12} = \frac{11}{12} .$ Answer: $\frac{11}{12}$ .

Recall Solution 3.2

Each answer is right with probability $\frac{1}{2}$ , independently. Let $X$ = number correct. We want $P (X \geq 2)$ . The complement here is not a single case — it's $X = 0$ or $X = 1$ : $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) .$ Using binomial counting over $2^{4} = 16$ equally likely answer-patterns:

$P (X = 0)$ : all $4$ wrong → $1$ pattern → $\frac{1}{16}$ .
$P (X = 1)$ : exactly one right → $(1 4) = 4$ patterns → $\frac{4}{16}$ . $P (X \geq 2) = 1 - \frac{1}{16} - \frac{4}{16} = 1 - \frac{5}{16} = \frac{11}{16} .$ Answer: $\frac{11}{16}$ .

Recall Solution 3.3

"Neither a heart nor a face card" is the complement of " $H$ or $F$ ". By De Morgan's idea, $H^{'} \cap F^{'} = (H \cup F)^{'}$ (everything outside both circles in the Venn picture above). First find $P (H \cup F)$ with the addition rule: $P (H \cup F) = P (H) + P (F) - P (H \cap F) .$

$P (H) = \frac{13}{52}$ (13 hearts).
$P (F) = \frac{12}{52}$ (3 face cards × 4 suits).
$P (H \cap F) = \frac{3}{52}$ (J, Q, K of hearts — the overlap). $P (H \cup F) = \frac{13 + 12 - 3}{52} = \frac{22}{52} = \frac{11}{26} .$ Now complement: $P (H^{'} \cap F^{'}) = 1 - \frac{11}{26} = \frac{15}{26} .$ Answer: $\frac{15}{26}$ . (Sanity check: $52 - 22 = 30$ cards are neither, and $30/52 = 15/26$ . ✓)

Level 4 — Synthesis

Goal: build the complement inside a multi-stage problem.

Recall Solution 4.1

Complement $A^{'} =$ "no defectives" $=$ "all $50$ good". Each item is good with probability $1 - 0.02 = 0.98$ : $P (A^{'}) = (0.98)^{50} .$ $P (A) = 1 - (0.98)^{50} .$ Numerically $(0.98)^{50} \approx 0.3642$ , so $P (A) \approx 1 - 0.3642 = 0.6358$ . Answer: $1 - (0.98)^{50} \approx 0.636$ .

Recall Solution 4.2

"At least one solves" has complement "none solves", i.e. all three fail. Failure probabilities are $1 - 0.5 = 0.5$ , $1 - 0.6 = 0.4$ , $1 - 0.7 = 0.3$ . By independence, multiply: $P (none solves) = 0.5 \times 0.4 \times 0.3 = 0.06.$ $P (at least one) = 1 - 0.06 = 0.94.$ Answer: $0.94$ .

Recall Solution 4.3

Complement $A^{'} =$ "all four digits distinct". Count distinct-string probability by filling positions one at a time: $P (A^{'}) = \frac{10}{10} \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10} = \frac{5040}{10000} = 0.504.$ $P (at least one repeat) = 1 - 0.504 = 0.496.$ Answer: $0.496$ .

Level 5 — Mastery

Goal: invent the complement in a problem that hides it.

Recall Solution 5.1

"A head appears within $5$ flips" has a messy direct description (it could arrive on flip $1$ , $2$ , ..., or $5$ ). The complement is one clean event: "no head in any of the $5$ flips" = "all $5$ are tails". $P (all tails) = (\frac{1}{2})^{5} = \frac{1}{32} .$ $P (head within 5 flips) = 1 - \frac{1}{32} = \frac{31}{32} .$ Answer: $\frac{31}{32}$ .

Recall Solution 5.2

Complement $A^{'} =$ "no dead bulb" $=$ "all $3$ drawn are good". Total ways to choose $3$ from $8$ is $(3 8) = 56$ ; ways to choose $3$ good from the $5$ good is $(3 5) = 10$ : $P (A^{'}) = \frac{( 3 5 )}{( 3 8 )} = \frac{10}{56} = \frac{5}{28} .$ $P (at least one dead) = 1 - \frac{5}{28} = \frac{23}{28} .$ Answer: $\frac{23}{28}$ .

Recall Solution 5.3

Direct "at least two share" spans a tangle of overlapping cases. The complement is orderly: "all $23$ birthdays are different." Fill birthdays one student at a time, each must avoid the previous ones: $P (all different) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \dots \frac{343}{365} = \prod_{k = 0}^{22} \frac{365 - k}{365} .$ Numerically this product is $\approx 0.4927$ . $P (at least two share) = 1 - 0.4927 \approx 0.5073.$ Answer: $\approx 0.507$ — just over $50%$ , the surprising result. The complement made a scary problem into a single tidy product.

Connections

Complementary events — P(A') = 1 − P(A) — the rule these exercises drill.
1.3.01-Sample-space-and-events — every $A^{'}$ lives inside the sample space $S$ .
1.3.03-Addition-rule-for-probabilities — needed for the "neither...nor" problem (3.3).
1.3.09-Independent-events — powers like $0.9 5^{3}$ , $0.9 8^{50}$ rely on independence.
1.3.12-At-least-one-problems — the core application, Levels 2–5.
1.4.05-Binomial-probability — the counting behind Problem 3.2.