Probability & Sets

IB Mathematics · Grade 9 · 20 Questions

Score 0 / 20
Set Theory Probability Combinations
Part 1 — Set Theory
Union, intersection, complement, Venn diagrams, subsets & set notation.
Q 01 Sets ⚡ A∪B = "either or both"

Given \(A = \{1, 2, 3, 4\}\) and \(B = \{3, 4, 5, 6\}\), find \(A \cup B\).

Quick Tip \(A \cup B\) means all elements in A or B (no duplicates). Think: U = Union = United family.

Explanation

Answer: B. \(A \cup B\) combines every element from both sets — remove duplicates. \(\{1,2,3,4\} \cup \{3,4,5,6\} = \{1,2,3,4,5,6\}\). The elements 3 and 4 appear in both, but we list them only once.
Q 02 Sets ⚡ A∩B = "overlap only"

Using the same sets \(A = \{1,2,3,4\}\) and \(B = \{3,4,5,6\}\), what is \(A \cap B\)?

Careful — this is NOT the same as union!

Explanation

Answer: C. Intersection \(\cap\) = elements that appear in both sets. Only 3 and 4 are in both A and B. Think: the ∩ symbol looks like a bridge connecting two sets — only what's on both sides counts.
Q 03 Sets ⚡ A' = "everything outside A"

If \(\mathcal{U} = \{1,2,3,4,5,6,7,8\}\) and \(A = \{2,4,6,8\}\), find \(A'\) (the complement of A).

Explanation

Answer: C. The complement \(A'\) contains all elements in the universal set \(\mathcal{U}\) that are not in A. Remove \(\{2,4,6,8\}\) from \(\{1,2,3,4,5,6,7,8\}\) → remaining = \(\{1,3,5,7\}\). Remember: \(A \cup A' = \mathcal{U}\) always.
Q 04 Sets ⚡ n(A∪B) = n(A)+n(B)−n(A∩B)

In a class of 30 students, 18 study French, 15 study Spanish, and 8 study both. How many study at least one language?

Use the inclusion-exclusion principle. Don't double-count!

Formula \(n(F \cup S) = n(F) + n(S) - n(F \cap S) = 18 + 15 - 8\)

Explanation

Answer: B. \(n(F \cup S) = 18 + 15 - 8 = 25\). We subtract 8 because those students were counted in both the French group and the Spanish group. Without subtracting, we'd count them twice.
Q 05 Sets ⚡ A⊆B means every element of A is in B

Which of the following is true?

Explanation

Answer: B. \(\{3\} \subseteq \{1,2,3,4\}\) is true because 3 is indeed inside \(\{1,2,3,4\}\). For A: 1 is not in \(\{2,3,4\}\). For C: 5 is not in the set. For D: 1 is not in \(\{2,3\}\). A subset means every element of the left set must appear in the right set.
Venn Diagrams
Q 06 Venn ⚡ Venn: draw first, count after

In a Venn diagram, region \((A \cap B)'\) represents:

Explanation

Answer: C. The prime ( ′ ) means "complement" = everything outside. \((A \cap B)'\) is the complement of the intersection, meaning all elements NOT in both A and B simultaneously. This includes: only-A elements, only-B elements, and elements outside both circles.
Part 2 — Probability
Part 2 — Probability
Basic probability, complementary events, combined events & conditional probability.
Q 07 Probability ⚡ P = favorable ÷ total

A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of picking a red marble?

Explanation

Answer: B and C are both correct. Total = 4+3+5 = 12. \(P(\text{red}) = \frac{4}{12} = \frac{1}{3}\). Both B and C equal \(\frac{1}{3}\). In this question, C is the unsimplified form. Always simplify your final answer unless told otherwise.
Q 08 Probability ⚡ P(A') = 1 − P(A)

The probability it rains tomorrow is 0.35. What is the probability it does NOT rain?

Complementary events always sum to 1.

Explanation

Answer: C. \(P(\text{no rain}) = 1 - P(\text{rain}) = 1 - 0.35 = 0.65\). The key rule: complementary events cover all possibilities, so they always add to exactly 1. D is impossible since probability can never exceed 1.
Q 09 Probability ⚡ Independent: P(A and B) = P(A)×P(B)

A coin is flipped and a die is rolled. What is the probability of getting a Head AND a 6?

Key Concept — Independent Events Two events are independent if one does not affect the other. Coin and die = independent. Multiply their probabilities.

Explanation

Answer: C. \(P(H) = \frac{1}{2},\ P(6) = \frac{1}{6}\). Since the coin and die are independent: \(P(H \cap 6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\).
Q 10 Probability ⚡ Mutually exclusive: P(A or B) = P(A)+P(B)

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

A card cannot be both a King and a Queen at the same time — these are mutually exclusive.

Explanation

Answer: C. 4 Kings + 4 Queens = 8 favorable cards out of 52. \(P = \frac{8}{52} = \frac{2}{13}\). Since a card can't be both simultaneously (mutually exclusive), we simply add: \(\frac{4}{52} + \frac{4}{52} = \frac{8}{52}\).
Q 11 Conditional ⚡ P(A|B) = P(A∩B) ÷ P(B)

In a class, 40% of students play football, 30% play basketball, and 15% play both. Given a student plays football, what is the probability they also play basketball?

This is conditional probability. The "|" symbol means "given that."

Formula \(P(B|F) = \dfrac{P(B \cap F)}{P(F)} = \dfrac{0.15}{0.40}\)

Explanation

Answer: B. \(P(B|F) = \frac{0.15}{0.40} = 0.375\). We narrow our sample space to only football players (40%). Among those, 15% also play basketball. So: \(\frac{15}{40} = \frac{3}{8} = 0.375\).
Q 12 Probability ⚡ Without replacement → denominator decreases

A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. What is \(P(\text{both red})\)?

Explanation

Answer: C. 1st draw: \(\frac{5}{8}\). After removing one red, 2nd draw: \(\frac{4}{7}\). Multiply: \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\). B also equals \(\frac{5}{14}\) — same answer, different form. Without replacement means the total decreases for each draw!
Part 3 — Combinations & Counting
Part 3 — Combinations & Counting
Factorial, permutations, combinations and the fundamental counting principle.
Q 13 Counting ⚡ n! = n×(n−1)×…×1

Evaluate \(5!\)

Factorial = multiply all positive integers down to 1. Used in permutations & combinations.

Explanation

Answer: C. \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Memorize common factorials: \(3!=6,\ 4!=24,\ 5!=120,\ 6!=720\). Note: \(0! = 1\) by definition!
Q 14 Permutation ⚡ Permutation = ORDER matters → P(n,r) = n!/(n−r)!

How many ways can 3 students be chosen and arranged in a line from a group of 8?

Order matters here — 1st, 2nd, 3rd place are different!

Formula \(P(n, r) = \dfrac{n!}{(n-r)!} = \dfrac{8!}{5!}\)

Explanation

Answer: B. \(P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\). Shortcut: start at 8 and multiply downward for 3 factors: \(8 \times 7 \times 6 = 336\).
Q 15 Combination ⚡ Combination = ORDER doesn't matter → C(n,r) = n!/(r!(n−r)!)

How many ways can a committee of 3 be selected from 8 people?

A committee has no rank — selecting Alice, Bob, Carol is the same as Carol, Bob, Alice.

Explanation

Answer: B. \(\binom{8}{3} = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\). Notice: \(C(8,3) = \frac{P(8,3)}{3!} = \frac{336}{6} = 56\). We divide by \(r!\) to remove the counted arrangements of the same group.
Q 16 Combination ⚡ C(n,r) = C(n, n−r) ← symmetry!

Which of the following equals \(\dbinom{10}{7}\)?

Hint: use the symmetry property of combinations.

Explanation

Answer: B. \(\binom{n}{r} = \binom{n}{n-r}\). So \(\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3}\). Both equal 120. This symmetry makes sense: choosing 7 to include is the same as choosing 3 to exclude.
Q 17 Counting ⚡ Fundamental Counting: multiply choices at each step

A restaurant offers 4 starters, 5 mains, and 3 desserts. How many different 3-course meals are possible?

Explanation

Answer: B. Fundamental Counting Principle: multiply. \(4 \times 5 \times 3 = 60\). Each choice is independent, so we multiply all possibilities together.
Q 18 Combination ⚡ Prob with combinations = C(favorable)/C(total)

A box has 6 good items and 4 defective items. 2 items are selected at random. What is the probability that both are good?

Strategy \(P = \dfrac{\binom{6}{2}}{\binom{10}{2}}\) — use combinations when order doesn't matter in selection.

Explanation

Answer: A. \(\binom{6}{2} = 15\), \(\binom{10}{2} = 45\). \(P = \frac{15}{45} = \frac{1}{3}\). Note: C also equals \(\frac{1}{3}\) — same value, just not simplified. Always reduce to lowest terms.
Q 19 Combined ⚡ Tree diagram: list ALL branches, sum = 1

Two dice are rolled. What is the probability that the sum equals 7?

Think systematically — list all pairs that sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Explanation

Answer: A. Total outcomes = \(6 \times 6 = 36\). Favorable outcomes (sum = 7): (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 outcomes. \(P = \frac{6}{36} = \frac{1}{6}\). Sum of 7 is actually the most likely sum on two dice!
Q 20 Challenge ⚡ P vs C: does order matter? YES→P, NO→C

A password consists of 2 letters (A–Z, no repeat) followed by 3 digits (0–9, no repeat). How many passwords are possible?

Order matters for passwords! Use permutations.

Explanation

Answer: A. Letters: first slot = 26 choices, second = 25 (no repeat). Digits: first = 10, second = 9, third = 8. Total = \(26 \times 25 \times 10 \times 9 \times 8 = 468{,}000\). B is wrong because combinations don't account for order — "AB" and "BA" are different passwords. C ignores the "no repeat" rule.

Your Final Score

Well done! 🎓

Review the explanations above to strengthen any weak areas.