Probability basics — sample space, events, P(E) = favourable - total: Exercises

This page is a self-test. Read each problem, try it yourself, THEN open the collapsible solution. Everything here is built on the parent topic — the sample space $S$ , an event $E$ , and the formula

$P (E) = \frac{n ( E )}{n ( S )} (when every outcome is equally likely) .$

Reminders of the vocabulary you'll lean on:

Sample Space — the full set $S$ of everything that can happen.
Events and Set Theory — an event is a subset of $S$ ; "or" is union $\cup$ , "and" is intersection $\cap$ .
Complement of an Event — $E^{'}$ is "everything not in $E$ ", and $P (E^{'}) = 1 - P (E)$ .
Mutually Exclusive Events — two events with no shared outcome; you may add their counts safely.
Counting Principles — multiply choices to count combinations.

Level 1 — Recognition

You just have to name the pieces: build $S$ , build $E$ , and divide.

Exercise L1.1

A single fair die is rolled. What is the probability of rolling a number less than 3?

Recall Solution

Step 1 — sample space. A die has 6 faces: $S = {1, 2, 3, 4, 5, 6}, n (S) = 6.$ Every face is equally likely, so the formula applies.

Step 2 — event. "Less than 3" means strictly below 3, so $1$ and $2$ only: $E = {1, 2}, n (E) = 2.$ Note $3$ is not included — "less than 3" excludes 3 itself.

Step 3 — divide. $P (E) = \frac{n ( E )}{n ( S )} = \frac{2}{6} = \frac{1}{3} \approx 0.333.$

Answer: $\frac{1}{3}$ .

Exercise L1.2

A bag has 4 red, 3 green, and 2 yellow marbles. One marble is drawn at random. What is $P (green)$ ?

Recall Solution

Step 1 — total outcomes. Add up all marbles: $n (S) = 4 + 3 + 2 = 9.$ Each marble is one equally-likely outcome.

Step 2 — favourable. There are 3 green marbles: $n (E) = 3$ .

Step 3 — divide. $P (green) = \frac{3}{9} = \frac{1}{3} .$

Answer: $\frac{1}{3}$ .

Level 2 — Application

Now you must construct the sample space from a rule, then filter it.

Exercise L2.1

Two fair coins are tossed. What is the probability of getting at least one tail?

Recall Solution

Step 1 — sample space. First coin: H or T. For each, second coin: H or T. That's $2 \times 2 = 4$ equally likely outcomes: $S = {HH, HT, TH, TT}, n (S) = 4.$ $HT$ and $TH$ are different because they say which coin was the tail.

Step 2 — event. "At least one tail" = one or more T: $E = {HT, TH, TT}, n (E) = 3.$

Step 3 — divide. $P (E) = \frac{3}{4} = 0.75.$

Faster check with the complement. The opposite of "at least one tail" is "no tails at all" = ${HH}$ , with probability $\frac{1}{4}$ . So $P (at least one tail) = 1 - \frac{1}{4} = \frac{3}{4} . ✓$

Answer: $\frac{3}{4}$ .

Exercise L2.2

A card is drawn from a standard 52-card deck. What is the probability it is a red card OR a King?

Recall Solution

Step 1 — sample space. $n (S) = 52$ .

Step 2 — count each piece.

Red cards (hearts + diamonds): $26$ .
Kings (one per suit): $4$ .

Step 3 — remove the overlap. Some cards are both red and a King: the King of hearts and the King of diamonds — that's $2$ cards counted in both piles. To count the union without double-counting: $n (red or King) = 26 + 4 - 2 = 28.$ This is the inclusion–exclusion idea from Events and Set Theory: $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ .

Step 4 — divide. $P (red or King) = \frac{28}{52} = \frac{7}{13} \approx 0.538.$

Answer: $\frac{7}{13}$ .

Level 3 — Analysis

Here the shape of the sample space matters. Order, replacement, and the 6×6 dice grid all change the counts. See the figures.

Exercise L3.1

Two fair dice are rolled. What is the probability the sum is 8 or more?

Recall Solution

Step 1 — sample space. Each die shows 1–6, so ordered pairs $(a, b)$ give $n (S) = 6 \times 6 = 36.$ Look at the grid in the figure — every cell is one equally-likely outcome.

Figure — Probability basics — sample space, events, P(E) = favourable - total

Step 2 — count sums $\geq 8$ . Read the shaded cells along the diagonals:

Sum $8$ : $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ → $5$
Sum $9$ : $(3, 6), (4, 5), (5, 4), (6, 3)$ → $4$
Sum $10$ : $(4, 6), (5, 5), (6, 4)$ → $3$
Sum $11$ : $(5, 6), (6, 5)$ → $2$
Sum $12$ : $(6, 6)$ → $1$

Total favourable: $5 + 4 + 3 + 2 + 1 = 15$ , so $n (E) = 15$ .

Step 3 — divide. $P (sum \geq 8) = \frac{15}{36} = \frac{5}{12} \approx 0.417.$

Answer: $\frac{5}{12}$ .

Exercise L3.2

Two balls are drawn without replacement from a box holding ${R, B, G}$ (one red, one blue, one green). What is the probability that one of the two drawn is red?

Recall Solution

Step 1 — sample space, ordered. Draw a first ball (3 choices), then a second from the remaining 2: $n (S) = 3 \times 2 = 6.$ Listing ordered pairs (first, second): $S = {(R, B), (R, G), (B, R), (B, G), (G, R), (G, B)} .$ "Without replacement" means the first ball can't come back, so $(R, R)$ is impossible — that's why the second factor is 2, not 3.

Step 2 — event. "One of the two is red" = red appears in the pair: $E = {(R, B), (R, G), (B, R), (G, R)}, n (E) = 4.$ The only pairs without red are $(B, G)$ and $(G, B)$ .

Step 3 — divide. $P (E) = \frac{4}{6} = \frac{2}{3} \approx 0.667.$

Answer: $\frac{2}{3}$ .

Level 4 — Synthesis

Combine several tools at once: complement + counting + careful sample space.

Exercise L4.1

Three fair coins are tossed. What is the probability of getting at least one head?

Recall Solution

Step 1 — sample space. Each coin: 2 outcomes, three coins: $n (S) = 2 \times 2 \times 2 = 8.$

Step 2 — use the complement. Counting "at least one head" directly means 7 outcomes. Counting the opposite — "zero heads" — is one outcome: $TTT$ . So $P (no heads) = \frac{1}{8} .$

Step 3 — subtract from 1 (the Complement of an Event rule): $P (at least one head) = 1 - \frac{1}{8} = \frac{7}{8} = 0.875.$

Answer: $\frac{7}{8}$ .

Exercise L4.2

A number is chosen at random from ${1, 2, \dots, 20}$ . What is the probability it is divisible by 2 or by 3?

Recall Solution

Step 1 — sample space. $n (S) = 20$ .

Step 2 — count each divisibility set.

Divisible by 2: ${2, 4, \dots, 20}$ → $10$ numbers.
Divisible by 3: ${3, 6, 9, 12, 15, 18}$ → $6$ numbers.
Divisible by both (i.e. by 6): ${6, 12, 18}$ → $3$ numbers. These are the overlap.

Step 3 — inclusion–exclusion. $n (2 or 3) = 10 + 6 - 3 = 13.$

Step 4 — divide. $P = \frac{13}{20} = 0.65.$

Answer: $\frac{13}{20}$ .

Level 5 — Mastery

Multi-stage reasoning where a wrong sample space quietly breaks everything.

Exercise L5.1

Two dice are rolled. Given nothing else, find the probability that the product of the two numbers is even.

Recall Solution

Step 1 — sample space. $n (S) = 36$ ordered pairs.

Step 2 — complement is cleaner. A product is odd only when both dice are odd (odd × odd = odd; any even factor makes the product even). Odd faces are ${1, 3, 5}$ — 3 choices each: $n (both odd) = 3 \times 3 = 9.$ So $P (product odd) = \frac{9}{36} = \frac{1}{4} .$

Step 3 — complement. $P (product even) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75.$

Answer: $\frac{3}{4}$ .

Exercise L5.2

A committee of 2 people is chosen at random from 3 women ${W_{1}, W_{2}, W_{3}}$ and 2 men ${M_{1}, M_{2}}$ . What is the probability the committee has exactly one woman?

Recall Solution

Step 1 — sample space (unordered pairs). A committee doesn't care about order, so we count combinations of 2 from 5 people. Using Counting Principles: $n (S) = (2 5) = \frac{5 \times 4}{2} = 10.$ These 10 pairs are all equally likely.

Step 2 — favourable: exactly one woman. Choose 1 woman from 3 and 1 man from 2: $n (E) = (1 3) \times (1 2) = 3 \times 2 = 6.$

Step 3 — divide. $P (exactly one woman) = \frac{6}{10} = \frac{3}{5} = 0.6.$

Answer: $\frac{3}{5}$ .

Exercise L5.3

Two dice are rolled. What is the probability that the maximum of the two faces is exactly 4?

Recall Solution

Step 1 — sample space. $n (S) = 36$ .

Step 2 — decode "maximum is exactly 4". Both dice must be $\leq 4$ , and at least one must equal 4.

Both $\leq 4$ : pairs from ${1, 2, 3, 4}$ → $4 \times 4 = 16$ .
Both $\leq 3$ (no 4 appears): $3 \times 3 = 9$ .
So "max is exactly 4" = (both $\leq 4$ ) minus (both $\leq 3$ ): $16 - 9 = 7$ .

Listing to confirm: $(4, 1), (4, 2), (4, 3), (4, 4), (1, 4), (2, 4), (3, 4)$ → $7$ pairs. $n (E) = 7$ .

Step 3 — divide. $P (max = 4) = \frac{7}{36} \approx 0.194.$

Answer: $\frac{7}{36}$ .

Recall Quick self-check answer key

L1.1 $::: \frac{1}{3}$ L1.2 $::: \frac{1}{3}$ L2.1 $::: \frac{3}{4}$ L2.2 $::: \frac{7}{13}$ L3.1 $::: \frac{5}{12}$ L3.2 $::: \frac{2}{3}$ L4.1 $::: \frac{7}{8}$ L4.2 $::: \frac{13}{20}$ L5.1 $::: \frac{3}{4}$ L5.2 $::: \frac{3}{5}$ L5.3 $::: \frac{7}{36}$

Next steps: solidify the union rule in Events and Set Theory, the complement shortcut in Complement of an Event, and the axioms these all rest on in Axioms of Probability.