📝 Chapter 1: Counting & Sample Spaces

Assessment AS Learning — Practice Quiz

🔄 Not Graded — Unlimited Retakes

Purpose: Self-check your understanding after each topic. This quiz is not graded. Use results to identify areas needing review before the unit test. Topics covered: sample spaces, fundamental counting principle, mutually exclusive events, Venn diagrams, conditional/independent probability.

Score: 0 / 12

Topic 1.1 — Sample Spaces

Question 1

A fair coin is tossed twice. What is the size of the sample space?

Solution:

Each toss has 2 outcomes. By the multiplicative principle: \( 2 \times 2 = 4 \).
Sample space: \( S = \{HH, HT, TH, TT\} \), so \( n(S) = 4 \).

Question 2

A standard die is rolled once. What is the probability of rolling a number less than 5?

\( \tfrac{1}{6} \)
\( \tfrac{2}{3} \)
\( \tfrac{5}{6} \)
\( \tfrac{1}{2} \)

Solution:

Favourable outcomes: \( \{1,2,3,4\} \), so 4 outcomes. Sample space: 6.
\( P(X<5) = \dfrac{4}{6} = \dfrac{2}{3} \).

Topic 1.2 — Fundamental Counting Principle

Question 3

A 4-digit PIN can use any digit 0–9. How many PINs are possible if digits may repeat?

Solution:

Each of 4 positions has 10 choices: \( 10 \times 10 \times 10 \times 10 = 10^4 = 10\,000 \).

Question 4

A licence plate has 3 letters (A–Z) followed by 3 digits (0–9), no repeats allowed. How many plates are possible?

Solution:

Letters (no repeat): \( 26 \times 25 \times 24 = 15\,600 \).
Digits (no repeat): \( 10 \times 9 \times 8 = 720 \).
Total: \( 15\,600 \times 720 = 11\,232\,000 \).

Topic 1.3 — Mutually Exclusive Events

Question 5

A card is drawn from a standard 52-card deck. What is \( P(\text{King or Queen}) \)?

\( \tfrac{1}{13} \)
\( \tfrac{2}{13} \)
\( \tfrac{4}{13} \)
\( \tfrac{1}{52} \)

Solution:

Kings and Queens are mutually exclusive (a card cannot be both).
\( P(K \cup Q) = P(K) + P(Q) = \dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13} \).

Question 6

A card is drawn from a standard deck. What is \( P(\text{Heart or Face card}) \)?

\( \tfrac{1}{4} \)
\( \tfrac{3}{13} \)
\( \tfrac{11}{26} \)
\( \tfrac{25}{52} \)

Solution:

Events are NOT mutually exclusive — there are 3 face cards that are also Hearts (J, Q, K of Hearts).
\( P(H \cup F) = P(H) + P(F) - P(H \cap F) = \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} = \dfrac{22}{52} = \dfrac{11}{26} \).

Topic 1.4 — Venn Diagrams

Question 7

In a class of 30 students, 18 study French and 14 study Spanish. 6 study both. How many study neither?

Solution:

\( n(F \cup S) = n(F) + n(S) - n(F \cap S) = 18 + 14 - 6 = 26 \).
Neither: \( 30 - 26 = 4 \).

Question 8

In a survey of 100 people: 60 like coffee, 50 like tea, 30 like both. How many like coffee OR tea?

Solution:

Inclusion–exclusion: \( |C \cup T| = |C| + |T| - |C \cap T| = 60 + 50 - 30 = 80 \).

Topic 1.5 — Independent & Dependent Events

Question 9

Two cards are drawn WITHOUT replacement from a standard deck. What is \( P(\text{both Aces}) \)?

\( \tfrac{4}{52} \cdot \tfrac{4}{52} = \tfrac{1}{169} \)
\( \tfrac{4}{52} \cdot \tfrac{3}{51} = \tfrac{1}{221} \)
\( \tfrac{4}{52} + \tfrac{4}{52} = \tfrac{2}{13} \)
\( \tfrac{1}{4} \)

Solution:

Without replacement, events are dependent. \( P(A_1 \cap A_2) = P(A_1) \cdot P(A_2 \mid A_1) \).
\( = \dfrac{4}{52} \cdot \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221} \).

Question 10

A bag has 5 red and 3 blue marbles. Two are drawn WITH replacement. \( P(\text{red, then blue}) \) = ?

\( \tfrac{15}{56} \)
\( \tfrac{15}{64} \)
\( \tfrac{8}{64} \)
\( \tfrac{1}{2} \)

Solution:

With replacement, events are independent.
\( P(R) \cdot P(B) = \dfrac{5}{8} \cdot \dfrac{3}{8} = \dfrac{15}{64} \).

Mixed Application

Question 11

A breakfast menu has 3 cereals, 4 fruits, and 2 drinks. How many different breakfasts (one of each) are possible?

Solution:

\( 3 \times 4 \times 2 = 24 \) breakfasts.

Question 12

A jar contains 4 red, 6 green, and 2 yellow candies. One is selected at random. \( P(\text{not green}) \) = ?

\( \tfrac{1}{2} \)
\( \tfrac{1}{4} \)
\( \tfrac{6}{12} \)
\( \tfrac{2}{3} \)

Solution:

Total: 12. Not green: \( 4 + 2 = 6 \).
\( P(\text{not green}) = \dfrac{6}{12} = \dfrac{1}{2} \). (Or: \( 1 - \dfrac{6}{12} = \dfrac{1}{2} \).)

📊 Self-Reflection

Rate your confidence on each topic (1 = need help, 4 = mastered):

1.1 Sample Spaces

1.2 Fundamental Counting Principle

1.3 Mutually Exclusive Events

1.4 Venn Diagrams

1.5 Independent & Dependent Events

Topics rated 1 or 2: Review the corresponding video lectures and retake those questions.