📝 Chapter 1: Counting & Sample Spaces

Assessment OF Learning — Unit Test
✅ Graded — Counts Toward 70% Term Mark
⏱️ Duration: 75 minutes  |  Total: /60 marks
Show all work. Answers without supporting work will receive partial credit at best.
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15 marks]
Question 1 [2 marks]
A coin is tossed 3 times. How many outcomes are in the sample space?
Solution:
\( 2 \times 2 \times 2 = 2^3 = 8 \) outcomes.
Question 2 [2 marks]
Two events \( A \) and \( B \) are mutually exclusive if:
Solution:
Mutually exclusive ⇔ disjoint ⇔ \( A \cap B = \varnothing \). Then \( P(A \cap B) = 0 \).
Question 3 [2 marks]
A meal is built by choosing 1 of 3 appetizers, 1 of 5 mains, and 1 of 4 desserts. How many possible meals?
Solution:
By the fundamental counting principle: \( 3 \times 5 \times 4 = 60 \) meals.
Question 4 [2 marks]
A card is drawn from a 52-card deck. \( P(\text{Spade or Ace}) = ? \)
Solution:
\( P(S \cup A) = \tfrac{13}{52} + \tfrac{4}{52} - \tfrac{1}{52} = \tfrac{16}{52} = \tfrac{4}{13} \) (Ace of Spades counted twice).
Question 5 [2 marks]
In a school, 60% take Math, 40% take Science, 25% take both. The percentage taking neither is:
Solution:
\( P(M \cup S) = 0.60 + 0.40 - 0.25 = 0.75 \). Neither: \( 1 - 0.75 = 0.25 = 25\% \).
Question 6 [2 marks]
Two cards are drawn without replacement from a 52-card deck. \( P(\text{both Hearts}) = ? \)
Solution:
Without replacement, dependent events: \( \tfrac{13}{52}\cdot\tfrac{12}{51} = \tfrac{156}{2652} = \tfrac{1}{17} \).
Question 7 [3 marks]
A 3-letter code uses letters from \( \{A,B,C,D,E\} \) without repetition. How many codes are possible?
Solution:
\( 5 \times 4 \times 3 = 60 \) codes.
Part B: Thinking [15 marks]
Question 8 [4 marks]
In a class of 100 students, 50 play hockey, 60 play soccer, and 30 play both. Use a Venn diagram to determine how many play exactly one of the two sports.
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Question 9 [4 marks]
A bag contains 6 red, 4 blue, and 2 green marbles. Two marbles are drawn without replacement. What is the probability that exactly one is red? Show your work using a tree diagram or case analysis.
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Question 10 [4 marks]
A 4-digit code uses digits 0–9. How many codes are even (last digit is 0, 2, 4, 6, or 8) AND have all distinct digits?
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Question 11 [3 marks]
If \( P(A) = 0.4 \), \( P(B) = 0.5 \), and \( P(A \cap B) = 0.2 \), determine if \( A \) and \( B \) are independent. Justify your answer using the test for independence.
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Part C: Communication [15 marks]
Question 12 [4 marks]
A student says: "If \( P(A) = 0.6 \) and \( P(B) = 0.5 \), then \( P(A \cup B) = 1.1 \), but probabilities can't exceed 1, so this is impossible." What is wrong with the student's reasoning? Provide a correct method.
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Question 13 [4 marks]
Explain the difference between mutually exclusive events and independent events. Give an example of each. Can two events be both mutually exclusive AND independent? Justify.
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Question 14 [4 marks]
Describe a step-by-step procedure for solving a "without replacement" probability problem. Include: identifying dependence, the formula, and the role of conditional probability.
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Question 15 [3 marks]
Compare a tree diagram and a Venn diagram. When is each more useful? Give a small example for each.
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Part D: Application [15 marks]
Question 16 [4 marks]
An online retailer's password requires 6 characters: the first must be a letter (A–Z, case-insensitive), the rest may be letters or digits, with repetition allowed. How many passwords are possible?
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Question 17 [4 marks]
In a survey of 200 commuters: 120 take the bus, 90 take the subway, 50 take both. a) How many take the bus only? b) How many take neither? c) If a commuter is selected at random, find \( P(\text{bus only}) \).
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Question 18 [3 marks]
In a quality-control test, 5% of widgets are defective. A widget is selected and tested. The test correctly identifies a defective widget 90% of the time and a good widget 95% of the time. Use a tree diagram to find \( P(\text{test reports defective}) \).
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Question 19 [4 marks]
A medical screening test has \( P(\text{positive} \mid \text{disease}) = 0.99 \) and \( P(\text{positive} \mid \text{no disease}) = 0.05 \). The disease prevalence is 1%. If a randomly chosen person tests positive, what is \( P(\text{disease} \mid \text{positive}) \)? Use a tree diagram.
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Evaluation Rubric

LevelDescription%
4Thorough, insightful, high degree of effectiveness80–100%
3Considerable effectiveness (provincial standard)70–79%
2Some effectiveness, approaching standard60–69%
1Limited effectiveness50–59%
RInsufficient achievementBelow 50%