Sample spaces, events, and axioms of probability: Exercises

Quick symbol reminder (all from the parent):

$Ω$ = the set of all possible outcomes (the "sample space").
$E \subseteq Ω$ = an event, a chosen subset of outcomes.
$∣ E ∣$ = how many outcomes are in $E$ (its size / count).
$P (\cdot)$ = the probability measure — a rule $P : events \to [0, 1]$ that takes an event and returns a number between 0 and 1 (how likely it is). It obeys the three axioms: $P (E) \geq 0$ , $P (Ω) = 1$ , and additivity over disjoint events. So $P (E)$ is read "the probability of event $E$ ."
$A \cup B$ = " $A$ or $B$ " (union), $A \cap B$ = " $A$ and $B$ " (intersection), $A^{c}$ = "not $A$ " (complement), $\emptyset$ = the impossible event (empty set).

The two-dice picture below is the workhorse for many problems — glance at it now.

Figure — Sample spaces, events, and axioms of probability

Each of the 36 cells is one equally-likely outcome $(die_{1}, die_{2})$ . Events are just regions you shade on this grid.

Level 1 — Recognition

Exercise 1.1 (L1)

A single fair coin is flipped twice. Write the sample space $Ω$ and its size $∣Ω∣$ . Then write the event $E =$ "at least one head."

Recall Solution

Each flip has outcomes $H$ or $T$ , and the two flips are ordered (first, second), so: $Ω = {H H, H T, T H, T T}, ∣Ω∣ = 4$ "At least one head" means not both tails, i.e. every outcome except $T T$ : $E = {H H, H T, T H}, ∣ E ∣ = 3$

Exercise 1.2 (L1)

For the two-dice experiment ( $∣Ω∣ = 36$ ), classify each as a simple event (one outcome), a compound event (many outcomes), the certain event, or the impossible event: (a) "both dice show 3" (b) "the sum is at least 2" (c) "the sum equals 1" (d) "the sum is even".

Recall Solution

(a) ${(3, 3)}$ — one outcome → simple event. (b) The smallest possible sum is $1 + 1 = 2$ , so every outcome qualifies → this is $Ω$ → certain event, $P = 1$ . (c) No two dice sum to $1$ → $\emptyset$ → impossible event, $P = 0$ . (d) Many outcomes (e.g. $(1, 1), (1, 3), (2, 2), \dots$ ) → compound event.

Level 2 — Application

Exercise 2.1 (L2)

Roll two fair dice. Compute $P (sum = 5)$ .

Recall Solution

Favourable pairs summing to 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ , so $∣ E ∣ = 4$ . Equally-likely outcomes → classical probability: $P (sum = 5) = \frac{∣ E ∣}{∣Ω∣} = \frac{4}{36} = \frac{1}{9} \approx 0.111$

Exercise 2.2 (L2)

A spam filter is tested on 2000 emails: 640 are spam, the rest not. Using the frequentist estimate, find $P (spam)$ and $P (not-spam)$ , and verify they satisfy Axiom 2.

Recall Solution

$P (spam) \approx \frac{640}{2000} = 0.32, P (not-spam) \approx \frac{2000 - 640}{2000} = \frac{1360}{2000} = 0.68$ Check (Axiom 2 / complement rule): $0.32 + 0.68 = 1.0$ ✓. This ratio-of-counts is exactly the MLE for a Bernoulli parameter.

Exercise 2.3 (L2)

A card is drawn from a standard 52-card deck. Let $A =$ "card is a heart." Find $P (A)$ and $P (A^{c})$ using the complement rule.

Recall Solution

There are 13 hearts, so $P (A) = \frac{13}{52} = \frac{1}{4} = 0.25$ . By the complement rule $P (A^{c}) = 1 - P (A) = 1 - 0.25 = 0.75$ (there are $52 - 13 = 39$ non-hearts, and $39/52 = 0.75$ ✓).

Level 3 — Analysis

Exercise 3.1 (L3)

Two fair dice. Let $A =$ "sum is 7" and $B =$ "at least one die shows 6." Find $P (A)$ , $P (B)$ , $P (A \cap B)$ , and $P (A \cup B)$ .

Recall Solution

Look at the shaded regions in the figure below.

$A$ = the anti-diagonal, $∣ A ∣ = 6$ → $P (A) = 6/36 = 1/6$ .
$B$ = the bottom row and rightmost column; the corner $(6, 6)$ is shared, so $∣ B ∣ = 6 + 6 - 1 = 11$ → $P (B) = 11/36$ .
$A \cap B$ = pairs that are on the diagonal and contain a 6: $(1, 6)$ and $(6, 1)$ , so $∣ A \cap B ∣ = 2$ → $P (A \cap B) = 2/36 = 1/18$ .
Inclusion–exclusion (needed because $A$ and $B$ overlap): $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{6}{36} + \frac{11}{36} - \frac{2}{36} = \frac{15}{36} = \frac{5}{12} \approx 0.417$

Exercise 3.2 (L3)

Draw one card. Let $H =$ "heart," $F =$ "face card (J, Q, K)." Find $P (H \cup F)$ .

Recall Solution

$∣ H ∣ = 13$ , $∣ F ∣ = 12$ (3 faces $\times$ 4 suits). The overlap $H \cap F$ = face cards that are hearts = ${J ♡, Q ♡, K ♡}$ , so $∣ H \cap F ∣ = 3$ . $P (H \cup F) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} \approx 0.423$

Exercise 3.3 (L3)

Given $P (A) = 0.5$ , $P (B) = 0.4$ , and $P (A \cup B) = 0.7$ . Find $P (A \cap B)$ and $P (A^{c} \cap B^{c})$ .

Recall Solution

Rearrange inclusion–exclusion for the overlap: $P (A \cap B) = P (A) + P (B) - P (A \cup B) = 0.5 + 0.4 - 0.7 = 0.2$ "Neither $A$ nor $B$ " is the complement of " $A$ or $B$ " (De Morgan: $A^{c} \cap B^{c} = (A \cup B)^{c}$ ): $P (A^{c} \cap B^{c}) = 1 - P (A \cup B) = 1 - 0.7 = 0.3$

Level 4 — Synthesis

Exercise 4.1 (L4)

Prove the general upper bound (Boole's inequality for two events): for any events $A, B$ , $P (A \cup B) \leq P (A) + P (B) .$ Then say exactly when equality holds.

Recall Solution

Start from inclusion–exclusion. Recall it is not an axiom — it was derived in the parent note by splitting $A \cup B$ into the disjoint pieces $A$ and $B \cap A^{c}$ and applying Axiom 3 (additivity over disjoint events). Taking that result as given: $P (A \cup B) = P (A) + P (B) - P (A \cap B) .$ By Axiom 1, $P (A \cap B) \geq 0$ . Subtracting a non-negative number can only make the total smaller or equal: $P (A \cup B) = P (A) + P (B) - \geq 0 P (A \cap B) \leq P (A) + P (B) . ■$ Equality holds iff $P (A \cap B) = 0$ — e.g. when $A$ and $B$ are mutually exclusive ( $A \cap B = \emptyset$ ).

Exercise 4.2 (L4)

Three mutually exclusive, exhaustive classes partition $Ω$ : $C_{1}, C_{2}, C_{3}$ with $P (C_{1}) = 0.5$ , $P (C_{2}) = 0.3$ . A classifier can output only one class per input. Find $P (C_{3})$ , and $P (C_{1} \cup C_{3})$ .

Recall Solution

"Exhaustive + mutually exclusive" means $C_{1} \cup C_{2} \cup C_{3} = Ω$ with no overlaps, so by Axioms 2 and 3 their probabilities sum to 1: $P (C_{1}) + P (C_{2}) + P (C_{3}) = 1 ⟹ P (C_{3}) = 1 - 0.5 - 0.3 = 0.2$ Since $C_{1}$ and $C_{3}$ are disjoint, Axiom 3 applies directly (no subtraction): $P (C_{1} \cup C_{3}) = P (C_{1}) + P (C_{3}) = 0.5 + 0.2 = 0.7$ This is the setup behind a softmax over 3 classes — the outputs are forced to sum to 1.

Exercise 4.3 (L4)

Using only the three axioms, prove that $P (\emptyset) = 0$ .

Recall Solution

The impossible event $\emptyset$ is disjoint from itself: $\emptyset \cap \emptyset = \emptyset$ . Also $\emptyset \cup \emptyset = \emptyset$ . Apply Axiom 3 to these two disjoint copies: $P (\emptyset) = P (\emptyset \cup \emptyset) = P (\emptyset) + P (\emptyset) = 2 P (\emptyset) .$ So $P (\emptyset) = 2 P (\emptyset)$ , which forces $P (\emptyset) = 0$ . (And Axiom 1 already told us it can't be negative, so this is the only consistent value.) $■$

Level 5 — Mastery

Exercise 5.1 (L5)

Model + solve. A model, on each email, is either "confident" (event $K$ , prob $0.6$ ) or "unsure" (event $K^{c}$ ). Given confident, it classifies correctly with prob $0.95$ ; given unsure, correctly with prob $0.5$ . Let $R$ = "model is correct." Using the axioms and the law of total probability (a consequence of Axiom 3 over a partition), find $P (R)$ .

Recall Solution

The pair ${K, K^{c}}$ partitions $Ω$ (mutually exclusive, exhaustive). Split $R$ along this partition: $R = (R \cap K) \cup (R \cap K^{c}), disjoint pieces.$ By Axiom 3, $P (R) = P (R \cap K) + P (R \cap K^{c})$ . Each piece is a conditional weighted by the class: $P (R) = P (R ∣ K) P (K) + P (R ∣ K^{c}) P (K^{c})$ $= (0.95) (0.6) + (0.5) (0.4) = 0.57 + 0.20 = 0.77.$ So the model is correct 77% of the time overall.

Exercise 5.2 (L5)

Prove and sanity-check. Show that for any three events, $P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C) .$ Then verify numerically with the two-dice events $A =$ "die 1 is 6", $B =$ "die 2 is 6", $C =$ "sum is 12".

Recall Solution

Proof (fold into two events). Let $U = A \cup B$ . Apply two-event inclusion–exclusion: $P (U \cup C) = P (U) + P (C) - P (U \cap C) .$ Now $P (U) = P (A) + P (B) - P (A \cap B)$ . Distribute the intersection: $U \cap C = (A \cap C) \cup (B \cap C)$ , so $P (U \cap C) = P (A \cap C) + P (B \cap C) - P ((A \cap C) \cap (B \cap C)),$ and $(A \cap C) \cap (B \cap C) = A \cap B \cap C$ . Substituting all three expansions and collecting terms gives the stated formula. $■$

Numeric check. $∣ A ∣ = 6, ∣ B ∣ = 6, ∣ C ∣ = ∣ {(6, 6)} ∣ = 1$ , all over 36:

$P (A) = P (B) = 6/36$ , $P (C) = 1/36$ .
$A \cap B = {(6, 6)}$ , $A \cap C = {(6, 6)}$ , $B \cap C = {(6, 6)}$ , $A \cap B \cap C = {(6, 6)}$ → each $= 1/36$ . $P (A \cup B \cup C) = \frac{6 + 6 + 1}{36} - \frac{1 + 1 + 1}{36} + \frac{1}{36} = \frac{13 - 3 + 1}{36} = \frac{11}{36} .$ Direct count: $A \cup B$ = "at least one 6" = 11 cells, and $C \subseteq A \cup B$ , so the union is 11 cells → $11/36$ ✓.

Recall Self-test recap (one line each)

$P (sum = 5)$ with two fair dice ::: $4/36 = 1/9 \approx 0.111$ Why can't you use $∣ E ∣/∣Ω∣$ on the sums ${2, \dots, 12}$ ? ::: sums are not equally likely; only the 36 ordered pairs are symmetric Inclusion–exclusion for two events ::: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $P (A^{c} \cap B^{c})$ in terms of the union ::: $1 - P (A \cup B)$ (De Morgan + complement) $P (\emptyset)$ from the axioms ::: $0$ , since $P (\emptyset) = 2 P (\emptyset)$ Law of total probability over partition ${K, K^{c}}$ ::: $P (R) = P (R ∣ K) P (K) + P (R ∣ K^{c}) P (K^{c})$